MA1100 Tutorial Xiang Sun 12 Dept. Mathematics November 15, 2009 - - PowerPoint PPT Presentation

. . . . . .

MA1100 Tutorial

MA1100 Tutorial

Xiang Sun12

• Dept. Mathematics

November 15, 2009

1Email: xiangsun@nus.edu.sg 2Corrections are always welcome.

. . . . . .

MA1100 Tutorial Introduction Self-Introduction

Self-Introduction

Name Sun Xiang (in English) or 孙祥 (in Chinese) Second year Ph.D student in Dept. Mathematics Email xiangsun@nus.edu.sg Phone 9053 5550 Office S9a-02-03 MSN xiangsun.sunny@hotmail.com QQ 402197754

. . . . . .

MA1100 Tutorial Introduction Tutorial Introduction

Introduction

There are 10 tutorials; Take attendance: 2 point for full attendance, and pro-rated for partial attendance; Things to do before Tutorial: Read lecture notes and textbook; Try to work through the tutorial problems before attending tutorial classes. Things to do after Tutorial: Understand the problems in the tutorial sets; Read lecture notes and textbook again, and understand everything in them.

. . . . . .

MA1100 Tutorial Tutorial 1: Sets and Logic

Schedule of Today

Review concepts Tutorial: 1.6, 1.7, 1.14, 1.26, 1.35, 1.38, 2.4, 2.11, 2.13, 2.21 Additional material

. . . . . .

MA1100 Tutorial Tutorial 1: Sets and Logic Review

Sets I

Notations {x P U | p(x)}, x is a general element, p(x) is the condition in terms

• f x, U is the universal set;

Relations Subsets: A B if every element of A is an element of B; Equality: A = B if A B and B A; Proper subsets: A B and A B; Empty set: H; Operations Power set: the power set of A is the set of all subsets of A, P(A) = {S U | S A} Intersection: the intersection of A and B is the set of all elements that are in both A and B, A X B = {x P U | x P A and x P B} Union: the union of A and B is the set of all elements that are in A or in B, A Y B = {x P U | x P A or x P B} Complement: the complement of A is the set of all elements

• f U that are not in A, Ac = {x P U | x R A}

Relative complement: the relative complement of B w.r.t. A is the set of all elements that are in A but not in B

. . . . . .

MA1100 Tutorial Tutorial 1: Sets and Logic Review

Sets II

Indexed Collection of Sets A1, A2, A3, . . . are an indexed collection of sets, N is index set Intersection: ∩

nPN An = A1 X A2 X A3 X

Union: ∪

nPN = A1 Y A2 Y A3 Y

Partitions of Sets A is a non-empty set, and S is a collection of subsets of A. S is a partition of A if For each X P S, X H, i.e. each part has at least one element; For every X, Y P S, if X Y, then X X Y = H; The union of all elements in the collection S is equal to A. Cartesian Products of Sets The Cartesian product of A and B: A B = {(a, b) | a P A, b P B}, (a, b) is an ordered pair

. . . . . .

MA1100 Tutorial Tutorial 1: Sets and Logic Review

Logic

Statements A statement is a sentence that is either true or false (but not both); We denote a statement by capital letters, usually P, Q, R, . . .; Open Sentences An open sentence is a (mathematical) sentence that involves variables; We denote an open sentence by capital letters with the variables involved, such as P(n), Q(x, y) Logic operators Let P and Q be two statements, Conjunction P ^ Q, it is true only when both P and Q are true; Disjunction P _ Q, it is false only when both P and Q are false; Negating P, if P is true, then P is false, and vice versa; Implication P Q, the rule is false only when the rule is violated. Necessary and Sufficient If S happens, then T happens, then S is sufficient for T, and T is necessary for S.

. . . . . .

MA1100 Tutorial Tutorial 1: Sets and Logic Tutorial

Exercise (1.6) The set E = {2x : x P Z} can be described by listing its element, namely E = {. . . , 4, 2, 0, 2, 4, . . .}. List the elements of the following sets in a similar manner. (a) A = {2x + 1 : x P Z} (b) B = {4n : n P Z} (c) C = {3q + 1 : q P Z} Recall There are two ways to describe a set: enumerating and describing. E1 = {2x : x P Z}–describing; E2 = {. . . , 4, 2, 0, 2, 4, . . .}–enumerating. Solution. For item (c), Z = {. . . , 2, 1, 0, 1, 2, . . .}, q

• 2
• 1

1 2 3q + 1

• 5
• 2

1 4 7 Therefore, We get that C can be listed as {. . . , 5, 2, 1, 4, 7, . . .}. Similarly, A = {. . . , 3, 1, 1, 3, 5, . . .}, and B = {. . . , 8, 4, 0, 4, 8, . . .}.

. . . . . .

MA1100 Tutorial Tutorial 1: Sets and Logic Tutorial

Exercise (1.7) The set E = {. . . , 4, 2, 0, 2, 4, . . .} of even integers can be described by means of a defining condition by E = {y = 2x : x P Z} = {2x : x P Z}. Describe the following sets in a similar manner. (a) A = {. . . , 4, 1, 2, 5, 8, . . .} (b) B = {. . . , 10, 5, 0, 5, 10, . . .} (c) C = {1, 8, 27, 64, 125, . . .} Method Find the rule of the sequences: For example, 1, 2, 5, 8, , . . .. It is an arithmetic sequence. Solution. A = {3x + 2 : x P Z} B = {5y : y P Z} C = {z3 : z P N}

. . . . . .

MA1100 Tutorial Tutorial 1: Sets and Logic Tutorial

Exercise (1.14) Find P ( P({1}) ) and its cardinality. Recall Power set: the power set of A is the set of all subsets of A, P(A) = {S U | S A}; Cardinality: the number of elements of set, denote as |S|. Method List all subsets. Solution. For {1}, its all subsets are H and {1}, then P({1}) = { H, {1} } ; For { H, {1} } , its all subsets are H, {H}, {{1}}, and { H, {1} } , therefore P ( P({1}) ) = { H, {H}, {{1}}, {H, {1}} } , and |P ( P({1}) ) | = 4

. . . . . .

MA1100 Tutorial Tutorial 1: Sets and Logic Tutorial

Exercise (1.26) For a real number r, define Ar = {r2}, Br as the closed interval [r 1, r + 1], and Cr as the interval (r, 8). For S = {1, 2, 4}, determine (a) ∪

αPS Aα and ∩ αPS Aα

(b) ∪

αPS Bα and ∩ αPS Bα

(c) ∪

αPS Cα and ∩ αPS Cα

Solution. For (a): Since S = {1, 2, 4}, there are three sets A1, A2, A4. By assumption, we have A1 = {1}, A2 = {4}, A4 = {16}. By definition, ∪

αPS

Aα = A1 Y A2 Y A4 = {1, 4, 16}, ∩

αPS

Aα = A1 X A2 X A4 = H. Similarly, B1 = [0, 2], B2 = [1, 3], B4 = [3, 5], and ∪

αPS

Bα = [0, 5], ∩

αPS

Bα = H. C1 = (1, 8), C2 = (2, 8), C4 = (4, 8), and ∪

αPS

Cα = (1, 8), ∩

αPS

Cα = (4, 8).

. . . . . .

MA1100 Tutorial Tutorial 1: Sets and Logic Tutorial

Exercise (1.35) Give an example of a set A with |A| = 4 and two disjoint partitions S1 and S2 of A with |S1| = |S2| = 3. Recall Partition

. . Click here

Solution. .

.

. 1 Let A = {a, b, c, d}, a, b, c, d are different. Now we want to seek some partition S consisting of 3 subsets; .

.

. 2 There are 6 partitions each of which consists of 3 subsets: {{a, b}, {c}, {d}}, {{a, c}, {b}, {d}}, {{a, d}, {b}, {c}}, {{b, c}, {a}, {d}}, {{b, d}, {a}, {c}}, {{c, d}, {a}, {b}}; .

.

. 3 Then just choose 2 of them, such that they are disjoint, for example {{a, b}, {c}, {d}} and {{c, d}, {a}, {b}}.

. . . . . .

MA1100 Tutorial Tutorial 1: Sets and Logic Tutorial

Exercise (1.38) Give an example of a partition of N into three subsets. Solution. Similar with the former exercise. There are many partitions, such as {{1}, {2}, {3, 4, . . .}}.

. . . . . .

MA1100 Tutorial Tutorial 1: Sets and Logic Tutorial

Exercise (2.4) The following is an open sentence over the domain R: P(x) : x(x 1) = 6. (a) For what values of x is P(x) a true statement? (b) For what values of x is P(x) a false statement? Solution. P(x) is true iff x satisfies the equation x(x 1) = 6. Solving the equation, we find x = 3 or x = 2, that is, P(x) is true iff x = 3 or x = 2. (a) When x = 3 or x = 2, P(x) is a true statement; (b) When x 3 and x 2, P(x) is a false statement;

. . . . . .

MA1100 Tutorial Tutorial 1: Sets and Logic Tutorial

Exercise (2.11) For the sets A = {1, 2, . . . , 10} and B = {2, 4, 6, 9, 12, 25}, consider the statements: P : A B, Q : |A B| = 6. Determine which of the following statements are true: (a) P _ Q; (b) P _ ( Q); (c) P ^ Q; (d) ( P) ^ Q; (e) ( P) _ ( Q). Recall Logic Operators

. . Click here

Method Use definitions. Solution. First, we get that P is false since 1 P A but 1 R B; Also we get that Q is true since A B = {1, 3, 5, 7, 8, 10}. Then we have (a) P _ Q is true; (b) P _ ( Q) is false; (c) P ^ Q is false; (d) ( P) ^ Q is true; (e) ( P) _ ( Q) is true;

. . . . . .

MA1100 Tutorial Tutorial 1: Sets and Logic Tutorial

Exercise (2.13) Let S = {1, 2, . . . , 6} and let P(A) : A X {2, 4, 6} = H and Q(A) : A H be open sentence over the domain P(S). (a) Determine all A P P(S) for which P(A) ^ Q(A) is true. (b) Determine all A P P(S) for which P(A) _ ( Q(A)) is true. (c) Determine all A P P(S) for which ( P(A)) ^ ( Q(A)) is true. Solution. (a) P(A) ^ Q(A) is true (and) { P(A) is true A X {2, 4, 6} = H Q(A) is true A H A must be a non-empty subset of {1, 3, 5}; (b) P(A) _ ( Q(A)) is true (or) { P(A) is true A X {2, 4, 6} = H Q(A) is false A = H A must be a subset of {1, 3, 5}. (c) ( P(A)) ^ ( Q(A)) is true (and) { P(A) is false A X {2, 4, 6} H Q(A) is false A = H

• contradiction, there is no such A.

. . . . . .

MA1100 Tutorial Tutorial 1: Sets and Logic Tutorial

Exercise (2.21) In each of the following, two open sentences P(x, y) and Q(x, y) are given, where the domain of both x and y is Z. Determine the truth value of P(x, y) Q(x, y) for the given values of x and y. (a) P(x, y) : x2 y2 = 0 and Q(x, y) : x = y. (x, y) P {(1, 1), (3, 4), (5, 5)}. (b) P(x, y) : |x| = |y| and Q(x, y) : x = y. (x, y) P {(1, 2), (2, 2), (6, 6)}. (c) P(x, y) : x2 + y2 = 1 and Q(x, y) : x + y = 1. (x, y) P {(1, 1), (3, 4), (0, 1), (1, 0)}. Solution. (a) When (x, y) = (1, 1), P(x, y) is true, Q(x, y) is false, then P(x, y) Q(x, y) is false; Similarly, truth for (x, y) = (3, 4) and (x, y) = (5, 5); (b) Truth for (x, y) = (1, 2) and (x, y) = (6, 6), false for (x, y) = (2, 2); (c) Truth for (x, y) = (1, 1), (x, y) = (3, 4) and (x, y) = (1, 0), false for (x, y) = (0, 1).

. . . . . .

MA1100 Tutorial Tutorial 1: Sets and Logic Additional material

Russell’s paradox

Exercise Usually, for any formal criterion, a set exists whose members are those objects (and

• nly those objects) that satisfy the criterion, i.e. {x P U : p(x)} is a set. Whether does

there exist an object with the form {x P U : p(x)}, which is not a set?

. . . . . .

MA1100 Tutorial Tutorial 1: Sets and Logic Additional material

Russell’s paradox

Solution. This question is disproved by a set containing exactly the sets that are not members of

• themselves. If such a set qualifies as a member of itself, it would contradict its own

definition as a set containing sets that are not members of themselves. On the other hand, if such a set is not a member of itself, it would qualify as a member of itself by the same definition. This contradiction is Russell’s paradox. Let A = {X P U : X R X}, U is the collection of all sets. If A P A, then A does not satisfy X R X, i.e. A R A, contradiction; If A R A, then A satisfies X R X, i.e. A P A, contradiction. Therefore, the definition is not well-defined, and the error is from U( U is not a set).

. . . . . .

MA1100 Tutorial Tutorial 1: Sets and Logic Additional material

Exercise How to solve this problem? Solution. Roughly speaking, the method is giving some restrictions on the definition of set. Russell’s paradox (also known as Russell’s antinomy), discovered by Bertrand Russell in 1901. In 1908, two ways of avoiding the paradox were proposed, Russell’s type theory and Ernst Zermelo’s axiomatic set theory, the first constructed axiomatic set theory. Zermelo’s axioms went well beyond Frege’s axioms of extensionality and unlimited set abstraction, and evolved into the now-canonical Zermelo-Fraenkel set theory (ZF). For more information, you can wiki “Russell’s paradox”.

. . . . . .

MA1100 Tutorial Tutorial 2: Logic

Schedule of Today

Review concepts Tutorial: 2.31(contradiction), 2.32(tautology), 2.37(equivalence), 2.40(negation), 2.48, 2.49(negation), 2.62(equivalence), 2.67, 2.68

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Review

Logic II(lecture 5)

Biconditional P if and only if Q, that is P Q. Converse Q P is called the converse of P Q. Contrapositive ( Q) ( P) is called the contrapositive of P Q. Tautology A logical expression that is always true is called a tautology. Contradiction A logical expression that is always false is called a contradiction. Logical Equivalence Two logical expressions are said to be logically equivalent to each

• ther if they have the same truth value.

De Morgan’s Law (P ^ Q) ( P) _ ( Q), (P _ Q) ( P) ^ ( Q). Distributive Law P_(Q^R) (P_Q)^(P_R), P^(Q_R) (P^Q)_(P^R). Implication as disjunction P Q ( P) _ Q. Negation of implication (P Q) P ^ ( Q). Implication with disjunction P (Q _ R) (P ^ ( Q)) R.

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Review

Logic III(lecture 6)

Universal quantifier The phrase “for each”, “for every”, “for all”, . . . is called a universal quantifier. Notation: @, say “for all”. Existential quantifier The phrase “there exists”, “there is”, . . . is called a existential

• quantifier. Notation: D, say “there exist”.

@ vs D P(x) true for (@x)P(x) (Dx)P(x) all the x True True

• nly some x

False True none of the x False False Two quantifiers (@x)(@y)P(x, y), (Dx)(Dy)P(x, y), (@y)(Dx)P(x, y), (Dx)(@y)P(x, y). Negation with quantifier (@x)P(x) (Dx)( P(x)), (Dx)P(x) (@x)( P(x)), (@x)(Dy)P(x, y) (Dx)(@y)( P(x, y)), (Dx)(@y)P(x, y) (@x)(Dy)( P(x, y)), (@x)(@y)P(x, y) (Dx)(Dy)( P(x, y)), (Dx)(Dy)P(x, y) (@x)(@y)( P(x, y))

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Tutorial

Exercise (2.31) For statements P and Q, show that ( P ^ ( Q) ) ^ (P ^ Q) is a contradiction. Recall A logical expression that is always false is called a contradiction. How to prove that a statement is contradiction? General method: It suffices to show that the statement is false for all combinations. Proof of Method 1. The compound statement ( P ^ ( Q) ) ^ (P ^ Q) is a contradiction since it is false for all combinations of truth values for the component statements P and Q. See the truth table below. P Q Q P ^ Q P ^ ( Q) ( P ^ ( Q) ) ^ (P ^ Q) T T F T F F T F T F T F F T F F F F F F T F F F

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Tutorial

Exercise (2.31) For statements P and Q, show that ( P ^ ( Q) ) ^ (P ^ Q) is a contradiction. Proof of Method 2. .

.

. 1 By associated law, we have ( P ^ ( Q) ) ^ (P ^ Q) P ^ ( Q) ^ P ^ Q. .

.

. 2 By commutative law, we have P ^ ( Q) ^ P ^ Q P ^ P ^ ( Q) ^ Q. .

.

. 3 Also by associated law, we have P ^ P ^ ( Q) ^ Q (P ^ P) ^ (Q ^ ( Q)) P ^ (Q ^ ( Q)). .

.

. 4 Since Q ^ ( Q) is always false, we have that P ^ (Q ^ ( Q)) is contradiction, i.e. ( P ^ ( Q) ) ^ (P ^ Q) is a contradiction.

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Tutorial

Exercise (2.32) For statements P and Q, show that ( P ^ (P Q) ) Q is a tautology. Then state ( P ^ (P Q) ) Q in words. (This is an important logical argument form, called modus ponens.) Recall A logical expression that is always true is called a tautology. How to prove that a statement is tautology? General method: It suffices to show that the statement is true for all combinations. Proof of Method 1. The compound statement ( P ^ (P Q) ) Q is a tautology since it is true for all combinations of truth values for the component statements P and Q. See the truth table below. P Q P Q P ^ (P Q) ( P ^ (P Q) ) Q T T T T T T F F F T F T T F T F F T F T ( P ^ (P Q) ) Q: If P and P implies Q, then Q.

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Tutorial

Exercise (2.32) For statements P and Q, show that ( P ^ (P Q) ) Q is a tautology. Then state ( P ^ (P Q) ) Q in words. (This is an important logical argument form, called modus ponens.) Proof of Method 2. .

.

. 1 Let R = P ^ (P Q), we want to show R Q is always true, that is, the cases in which R Q is false will not happen. .

.

. 2 R Q is false only when R is true and Q is false. If we prove that there is a contradiction, i.e., this case will not happen, then it is done. .

.

. 3 Suppose that R is true and Q is false. .

.

. 4 Since R = P ^ (P Q), we have that P and P Q are both true. .

.

. 5 We get that Q is true. It is a contradiction. Therefore, this case will not happen, that is, the statement is a tautology.

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Tutorial

Exercise (2.37) For statements P and Q, show that ( Q) (P ^ ( P)) and Q are logically equivalent. Recall Two logical statements are equivalent if they have the same truth value. How to prove that two statements are equivalent? General method: It suffices to show that two statements have same truth value. Proof of Method 1. The statements Q and ( Q) ( P ^ ( P) ) are logically equivalent since they have the same truth values for all combinations of truth values for the component statements P and Q. See the truth table below. P Q P Q P ^ ( P) ( Q) ( P ^ ( P) ) T T F F F T T F F T F F F T T F F T F F T T F F

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Tutorial

Exercise (2.37) For statements P and Q, show that ( Q) (P ^ ( P)) and Q are logically equivalent. Proof of Method 2. .

.

. 1 Since P ^ ( P) is always false, it is enough to consider Q. .

.

. 2 When Q is true, then ( Q) (P ^ ( P)) is true, the same as Q. .

.

. 3 When Q is false, then ( Q) (P ^ ( P)) is false, the same as Q.

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Tutorial

Exercise (2.40) Write negations of the following open sentences: (a) Either x = 0 or y = 0. (b) The integers a and b are both even. Recall De Morgan’s laws: (P ^ Q) ( P) _ ( Q), (P _ Q) ( P) ^ ( Q). Roughly speaking, we can consider as an operator, which changes ^(and) to _(or), and changes _(or) to ^(and). Solution. (a) Let P : x = 0 and Q : y = 0. Then the statement can be expressed as P _ Q. Therefore the negation is ( P) ^ ( Q), i.e. both x 0 and y 0. (b) Let P : a is even and Q : b is even, then the statement can be expressed as P ^ Q. Therefore the negation is ( P) _ ( Q), i.e. either a is odd or b is odd.

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Tutorial

Exercise (2.48) Determine the truth value of each of the following statements. (a) Dx P R, x2 x = 0. (b) @n P N, n + 1 2. (c) @x P R, √ x2 = x. (d) Dx P Q, 3x2 27 = 0. (e) Dx P R, Dy P R, x + y + 3 = 8. (f) @x, y P R, x + y + 3 = 8. (g) Dx, y P R, x2 + y2 = 9. (h) @x P R, @y P R, x2 + y2 = 9. Recall

True False (@x)P(x) P(x) is true for all x P(x) is false for some x (Dx)P(x) P(x) is true for some x P(x) is false for all x (@x)(@y)P(x, y) P(x, y) is true for all x and all y P(x, y) is false for some x or some y (Dx)(Dy)P(x, y) P(x, y) is true for some x and some y P(x, y) is false for all x and y (Dx)(@y)P(x, y) For some x, P(x, y) is true for all y For any x, P(x, y) is false for some y (@x)(Dy)P(x, y) For any x, P(x, y) is true for some y For some x, P(x, y) is false for all y

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Tutorial

Solution. (a) True. 0 and 1 are the solutions of x2 x = 0. (b) True. For any n P N, we have n 1 since N = {1, 2, 3, . . .}. Therefore n + 1 2. (c) False. Let x 0, then √ x2 = x x. (d) True. 3 and -3 are the solutions of 3x2 = 27. (e) True. x = 0, y = 5 is the solution of x + y + 3 = 8. (f) False. x = 0, y = 0 does not satisfy x + y + 3 = 8. (g) True. x = 0, y = 3 is the solution of x2 + y2 = 9. (h) False. x = 0, y = 0 does not satisfy x2 + y2 = 9.

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Tutorial

Exercise (2.49) The statement For every integer m, either m 1 or m2 4. can be expressed using a quantifier as: @m P Z, m 1 or m2 4. Do this for the statements in parts (a) and (b). (a) There exist integers a and b such that both ab 0 and a + b 0. (b) For all real numbers x and y, x y implies that x2 + y2 0. (c) Express in words the negations of the statements in (a) and (b). (d) Using quantifiers, express in symbols the negations of the statements in both (a) and (b).

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Tutorial

Recall Roughly speaking, changes @ to D, and changes D to @. Implication as disjunction: P Q ( P) _ Q. Negation of implication: (P Q) P ^ ( Q). Solution. (a) Da, b P Z, ab 0 and a + b 0. (b) @x, y P R, x y implies x2 + y2 0. (c-d) (a) @a, b P Z, either ab 0 or a + b 0, that is, for all integers a and b, either ab 0 or a + b 0. (b) Dx, y P R, x y and x2 + y2 0, that is, there exist real numbers x and y such that x y and x2 + y2 0. Besides, the negation of (b) can be: Dx, y P R, x y and x2 + y2 = 0 since x2 + y2 0 for all x, y P R.

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Tutorial

Exercise (2.62) (a) For statements P, Q, and R, show that ( (P ^ Q) R )

• (

(P ^ ( R)) ( Q) ) . (b) For statements P, Q, and R, show that ( (P ^ Q) R )

• (

(Q ^ ( R)) ( P) ) .

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Tutorial

Proof of Method 1. The logical expressions are said to be locally equivalent to each other if they have the same truth value.

Part(a): P Q R Q R P ^ Q P ^ ( R) ( (P ^ Q) R ) ( (P ^ ( R)) ( Q) ) T T T F F T F T T T F T T F F F T T F T T F F F F T T F F T T F F F T T T T F F T T T F F T F F T T F T T T F T F F T F F T T F F F T T F F T T Part(b): P Q R P R P ^ Q Q ^ ( R) ( (P ^ Q) R ) ( (Q ^ ( R)) ( P) ) T T T F F T F T T T F T F F F F T T F T T T F F F T T F F T T F F F T T T T F F T T T F F T F F F T F F T T F T F T T F T T T F F F T T F F T T

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Tutorial

Proof of Method 2. (a) (P ^ Q) R (P ^ Q) _ R by implication as disjunction ( P) _ ( Q) _ R by De Morgan’s law (P ^ ( R)) ( Q) (P ^ ( R)) _ ( Q) by implication as disjunction ( P) _ R _ ( Q) by De Morgan’s law ( P) _ ( Q) _ R by commutative law Therefore, we have ( (P ^ Q) R )

• (

(P ^ ( R)) ( Q) ) . (b) .

.

. 1 Let Q1 = P and P1 = Q, then the statement can be written as ( (Q1 ^ P1) R )

• (

(P1 ^ ( R)) ( Q1) ) . .

.

. 2 By commutative law, we have ( (P1 ^ Q1) R )

• (

(P1 ^ ( R)) ( Q1) ) . .

.

. 3 Apply part (a).

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Tutorial

Exercise (2.67) Do there exist a set S of cardinality 2 and a set {P(n), Q(n), R(n)} of three open sentences over the domain S such that the implications P(a) Q(a), Q(b) R(b), and R(c) P(c) are true, where a, b, c P S, and (2) the converses of the implications in (1) are false? Necessarily, at least two of these elements a, b. Solution. .

.

. 1 If Q(a) P(a), R(b) Q(b), and P(c) R(c) are false, then Q(a), R(b), P(c) must be true, and P(a), Q(b), R(c) must be false. .

.

. 2 Since |S| = 2, there are at least two of a, b, c which are same. .

.

. 3 If a = b, since Q(a) is true and Q(b) is false, it is a contradiction. .

.

. 4 If a = c, since P(c) is true and P(a) is false, it is a contradiction. .

.

. 5 If b = c, since R(b) is true and R(c) is false, it is a contradiction. .

.

. 6 Therefore, there does not exist such set S.

. . . . . .

MA1100 Tutorial Tutorial 2: Logic Tutorial

Exercise (2.68) Let A = {1, 2, . . . , 6} and B = {1, 2, . . . , 7}. For x P A, let P(x) : 7x + 4 is odd. For y P B, let Q(y) : 5y + 9 is odd. Let S = {( P(x), Q(y) ) : x P A, y P B, P(x) Q(y) is false } . What is |S|? Recall For given x0, P(x0) is a sentence, not a truth value. Solution. .

.

. 1 P(x) is true for x = 1, 3, 5 and false for x = 2, 4, 6. .

.

. 2 Q(y) is true for y = 2, 4, 6 and false for y = 1, 3, 5, 7. .

.

. 3 P(x) Q(y) is false if P(x) is true and Q(y) is false. Thus S = {(P(x), Q(y) : x = 1, 3, 5, y = 1, 3, 5, 7} = {P(x) : x = 1, 3, 5} {Q(y) : y = 1, 3, 5, 7}, and |S| = 3 4 = 12.

. . . . . .

MA1100 Tutorial Tutorial 3: Proof

Schedule of Today

Review concepts Tutorial: 3.10(parity), 3.18(parity, biconditional, direct proof, contrapositive), 3.20(by cases), 3.22(parity), 3.27, 3.28, 3.29, 4.4(congruent, by cases), 4.12(contrapositive, by cases), 4,15(congruent)

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Review

True statements and Proof

True statements Definition: Giving the precise meaning of a word or phase that represents some object, property or other concepts. Axiom: Basic properties that are regarded as true statement without needing a proof is called an axiom. Theorem, lemma, corollary, proposition (need proofs). Axioms, Definitions Theorems, Lemmas, Propositions. Proof Direct proof: Starting from hypothesis P, using some true statements to get conclusion Q. Proof by contrapositive: P Q ( Q) ( P). Proof by cases: for convenience, we usually split the assumption to several cases, then prove every case. Disjunction in conclusion: (P (Q _ R)) ((P ^ ( Q)) R). Advantage: more conditions. Proving biconditionals: P Q (P Q) ^ (Q P).

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Review

Integers

Basic properties of (Z, +, ): Identity: n + 0 = n and n 1 = n; Inverse: n + (n) = 0, but the inverse for multiplication does not exist except n = 1; Commutative: n + m = m + n and m n = n m; Associative: (m + l) + n = m + (l + n) and (m l) n = m (l n); Distributive: m (l + n) = m l + m n, and (m + l) n = m n + l n; Closure: Z is closed under { addition m + n multiplication m n P Z for any m, n P Z. Parity: n is {

• dd

even, iff there exists an integer m, such that n = { 2m + 1 2m . There are some facts:

• ddodd=even, oddeven=odd, eveneven=even. (By definition)

n is even iff n2 is even; (Theorem 3.12) n is odd iff n2 is odd; (Contrapositive of Theorem 3.12) ab is even iff a is even or b is even. (Theorem 3.17) ab is odd iff a is odd and b is odd. (Contrapositive of Theorem 3.17)

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Review

Integers

Divisibility: m divides n if there exists an integer q, such that n = mq. Notation: m n, and we say that m is divisor. Negation: m does not divide n if for any integer q, n mq. Notation: m n. Congruence: Let a, b and n be integers with n 1. If n divides a b, we say that a is congruent to b modulo n. Notation: a b (mod n). If a b (mod n), then a = b + nk for some integer k. Relation: Let a and n be integers with n 1. a 0 (mod n) iff n a. Division Algorithm3: Given two integers a and d, with d 0. There exist unique integers q and r such that a = qd + r and 0 r |d|, where |d| denotes the absolute value of d. q is called the quotient, r is called the remainder, d is called the divisor, and a is called the dividend. That is, for any integers a and d (here we assume that d is positive), we have that a can be expressed as a = qd, a = qd + 1, . . . , or a = qd + (d 1) for some integer q. For example, let d = 3, then every integer x can be expressed as x = 3q, x = 3q + 1, or x = 3q + 2 for some integer q.

3Ref Theorem 11.4 on page 247

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Review

Real Numbers

Absolute value: |x| = { x, if x 0; x, if x 0. Triangle inequality: |x + y| |x| + |y|. Proof. By definition, we have |x| x |x| and |y| y |y|, then (|x| + |y|) x + y |x| + |y|, also by definition, we have |x + y| |x| + |y|. Triangle inequality: |x y| |x| |y|.

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Exercise (3.10) Let x P Z. Prove that if 22xis an odd integer, then 4x is an odd integer. Proof. For any x P Z, we have 22x = ( 22)x = 4x. Therefore 4x is odd iff 22x is odd.

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Exercise (3.18) Let n P Z. Prove that (n + 1)2 1 is even if and only if n is even. Proof of Method 1. There are two directions to be proved: “If” Suppose that n is even, we want to show that (n + 1)2 1 is even. There are two methods to prove it: First we use a direct proof: Let n = 2k where k is an integer. Then (n + 1)2 1 = 4k2 + 4k = 2(2k2 + 2k) is even. Also we can use a proof by contrapositive: assume that (n + 1)2 1 is

• dd, we want to show that n is odd: (n + 1)2 1 = n(n + 2) is odd, then

both n and n + 2 are odd (by contrapositive of Thm 3.17). “Only if” Suppose that (n + 1)2 1 is even, we want to show that n is even. There are also two methods to prove it: First we use a direct proof: Since (n + 1)2 1 = n(n + 2) is even, we have that n is even or n + 2 is even. If n is even, okay, there is nothing to prove; if n + 2 is even, let n + 2 = 2k, then n = 2(k 1) is even. Also we can use a proof by contrapositive: assume that n is odd, we want to show that (n + 1)2 1 is odd: Let n = 2m + 1, then (n + 1)2 1 = 4m2 + 8m + 3 is odd.

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Exercise (3.18) Let n P Z. Prove that (n + 1)2 1 is even if and only if n is even. Proof of Method 2. .

.

. 1 We have (n + 1)2 1 = n2 + 2n + 1 1 = n2 + 2n. .

.

. 2 Since 2n is always even, (n + 1)2 1 is even if and only if n2 is even. .

.

. 3 n2 is even if and only if n is even. (By theorem 3.12)

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Exercise (3.20) Prove that if n P Z, then n3 n is even. Recall This question is about parity, then we will split its assumptions to some cases. Proof of Method 1. Let n P Z. We consider two cases. .

.

. 1 n is even. Then n = 2a for some integer a. Thus n3 n = 8a3 2a = 2(4a3 a). Since 4a3 a is an integer, n3 n is even. .

.

. 2 n is odd. Then n = 2b + 1 for some integer b. Observe that n3 n = (2b + 1)3 (2b + 1) = 8b3 + 12b2 + 6b + 1 2b 1 = 8b3 + 12b2 + 4b = 2(4b3 + 6b2 + 2b) Since 4b3 + 6b2 + 2b is an integer, n3 n is even.

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Exercise (3.20) Prove that if n P Z, then n3 n is even. Proof of Method 2. Let n P Z, then n3 n = n(n + 1)(n 1). If n is odd, then both n 1 and n + 1 are even, therefore n3 n is even (By Theorem 3.17); If n is even, then there is nothing to prove since eveninteger=even (By Theorem 3.17).

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Exercise (3.22) Let a, b P Z. Prove that if ab is odd, then a2 + b2 is even. Proof. By theorem 3.17: ab is even iff a is even or b is even, we have that ab is odd iff a is odd and b is odd (by contrapositive). By theorem 3.12: n2 is even iff n is even, we have that both a2 and b2 are odd (also by contrapositive). Since oddodd=even, we have that a2 + b2 is even.

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Exercise (3.27) Below is a proof of a result. Proof: We consider two cases. a and b are even. Then a = 2r and b = 2s for some integers r and s. Thus a2 b2 = (2r)2 (2s)2 = 4r2 4s2 = 2(2r2 2s2). Since 2r2 2s2 is an integer, a2 b2 is even. a and b are odd. Then a = 2r + 1 and b = 2s + 1 for some integers r and s. Thus a2 b2 = (2r + 1)2 (2s + 1)2 = (4r2 + 4r + 1) (4s2 + 4s + 1) = 4r2 + 4r 4s2 4s = 2(2r2 + 2r 2s2 2s). Since 2r2 + 2r 2s2 2s is an integer, a2 b2 is even. Which of the following is being proved? .

.

. 1 Let a, b P Z. Then a and b are of the same parity if and only if a2 b2 is even. .

.

. 2 Let a, b P Z. Then a2 b2 is even. .

.

. 3 Let a, b P Z. If a and b are of the same parity, then a2 b2 is even. .

.

. 4 Let a, b P Z. If a2 b2 is even, then a and b are of the same parity.

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Method For these questions, we focus on the assumptions and conclusions, and no need to check the proof. Solution. There is only one direction, then (1) is not proved. There are two cases: a, b are both even or a, b are both odd, then (2) is not proved. (3) is proved. Since (4) is the converse of (3), (4) is not proved.

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Exercise (3.28) Below is given a proof of a result. What result is being proved? Proof: Assume that x is even. Then x = 2a for some integer a. So 3x2 4x 5 = 3(2a)2 4(2a) 5 = 12a2 8a 5 = 2(6a2 4a 3) + 1. Since 6a2 4a 3 is an integer, 3x2 4x 5 is odd. For the converse, assume that x is odd. So x = 2b + 1, where b P Z. Therefore, 3x2 4x 5 = 3(2b + 1)2 4(2b + 1) 5 = 3(4b2 + 4b + 1) 8b 4 5 = 12b2 + 4b 6 = 2(6b2 + 2b 3). Since 6b2 + 2b 3 is an integer, 3x2 4x 5 is even.

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Solution. There are two parts of the proof: The first part: x is even 3x2 4x 5 is odd; The second part: x is odd 3x2 4x 5 is even; This statement is equivalent to its contrapositive: x is even 3x2 4x 5 is odd. Therefore, the result which has been proved is that for any integer x, x is even if and only if 3x2 4x 5 is odd. This can also be restated as: Let x P Z. Then x is odd if and only if 3x2 4x 5 is even.

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Exercise (3.29) Evaluate the proof of the following result. Result: Let n P Z. If 3n 8 is odd, then n is odd. Proof: Assume that n is odd. Then n = 2k + 1 for some integer k. Then 3n 8 = 3(2k + 1) 8 = 6k + 3 8 = 6k 5 = 2(3k 3) + 1. Since 3k 3 is an integer, 3n 8 is odd. Solution. From “3n 8 is odd”, we want to show that “n is odd”, but the proof shows its

• converse. No proof has been given of the result itself.

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Exercise (4.4) Let x, y P Z. Prove that if 3 x and 3 y, then 3 (x2 y2). Proof of Method 1. Assume that 3 x and 3 y. Then by division algorithm, we have that x = 3p + 1 or x = 3p + 2 for some integer p and y = 3q + 1 or y = 3q + 2 for some integer q. Then we consider the following four cases. .

.

. 1 x = 3p + 1 and y = 3q + 1. Then x2 y2 = (3p + 1)2 (3q + 1)2 = (9p2 + 6p + 1) (9q2 + 6q + 1) = 3(3p2 + 2p 3q2 2q). Since 3p2 + 2p 3q2 2q is an integer, 3 (x2 y2). Using similar arguments for the remaining cases. .

.

. 2 x = 3p + 1 and y = 3q + 2. .

.

. 3 x = 3p + 2 and y = 3q + 1. .

.

. 4 x = 3p + 2 and y = 3q + 2.

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Exercise (4.4) Let x, y P Z. Prove that if 3 x and 3 y, then 3 (x2 y2). Proof of Method 2. x2 y2 = (x + y)(x y), then it suffices to show that at least on of (x + y) and (x y) can be 3 divided (By result 4.3). Also consider the four cases before: x = 3p + 1 and y = 3q + 1, then 3 (x y). x = 3p + 1 and y = 3q + 2, then 3 (x + y). x = 3p + 2 and y = 3q + 1, then 3 (x + y). x = 3p + 2 and y = 3q + 2, then 3 (x y).

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Exercise (4.12) Let a, b P Z. Prove that if a2 + 2b2 0 (mod 3), then either a and b are both congruent to 0 modulo 3 or neither is congruent to 0 modulo 3. Method If the conclusion is complicated, we may consider to show its contrapositive. The

• riginal statement is (a2 + 2b2 0)

( (a, b 0) _ (a, b 0) ) . Then the contrapositive is (( (a, b 0) ) ^ ( (a, b 0) )) (a2 + 2b2 0). The left hand side is

( (a 0, b 0)_(a 0, b 0)_(a 0, b 0) ) ^ ( (a 0, b 0)_(a 0, b 0)_(a 0, b 0) ) .

Then by distributive law, left hand side is (a 0, b 0) _ (a 0, b 0).

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Method Then the contrapositive is ( (a 0, b 0) _ (a 0, b 0) ) (a2 + 2b2 0). Therefore, there are 2 cases, and each of which has two subcases:                a 0 (mod 3) and b 0 (mod 3) { a = 3p and b = 3q + 1 a = 3p and b = 3q + 2 a 0 (mod 3) and b 0 (mod 3) { a = 3p + 1 and b = 3q a = 3p + 2 and b = 3q

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Proof. Case 1 Assume that a 0 (mod 3) and b 0 (mod 3). .

.

. 1 b = 3q + 1. Then a2 + 2b2 = (3p)2 + 2(3q + 1)2 = 9p2 + 2(9q2 + 6q + 1) = 9p2 + 18q2 + 12q + 2 = 3(3p2 + 6q2 + 4q) + 2. Since 3p2 + 6q2 + 4q is an integer, 3 (a2 + 2b2) and so a2 + 2b2 0 (mod 3). .

.

. 2 b = 3q + 2. (The proof is similar to that of Subcase 1.1.) Case 2 Assume that a 0 (mod 3) and b 0 (mod 3). .

.

. 1 a = 3p + 1. Then a2 + 2b2 = (3p + 1)2 + 2(3q)2 = 9p2 + 6p + 1 + 18q2 = 9p2 + 6p + 18q2 + 1 = 3(3p2 + 2p + 6q2) + 1. Since 3p2 + 2p + 6q2 is an integer, 3 (a2 + 2b2) and so a2 + 2b2 0 (mod 3). .

.

. 2 a = 3p + 2. (The proof of each subcase is similar to that of Subcase 2.1.)

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Exercise (4.12) Let a, b P Z. Prove that if a2 + 2b2 0 (mod 3), then either a and b are both congruent to 0 modulo 3 or neither is congruent to 0 modulo 3. Method There is another method based on Disjunction in conclusion: (P (Q _ R)) ((P ^ ( Q)) R) . If we denote P : a2 + 2b2 0, Q : a 0, b 0, R : a 0, b 0, then the original statement is P (Q _ R). By Disjunction in conclusion: if we prove that P ^ ( Q) R, i.e. a2 + 2b2 0, a 0 or b 0 a 0, b 0, then it has been done.

. . . . . .

MA1100 Tutorial Tutorial 3: Proof Tutorial

Exercise (4.15) Let a, b P Z. Show that if a 5 (mod 6) and b 3 (mod 4), then 4a + 6b 6 (mod 8). Proof. Assume that a 5 (mod 6) and b 3 (mod 4). Then 6 (a 5) and 4 (b 3). Thus a 5 = 6x and b 3 = 4y, where x, y P Z. So a = 6x + 5 and b = 4y + 3. Observe that 4a + 6b = 4(6x + 5) + 6(4y + 3) = 24x + 20 + 24y + 18 = 24x + 24y + 38 = 8(3x + 3y + 4) + 6. Since 3x + 3y + 4 is an integer, 8 (4a + 6b 6) and so 4a + 6b 6 (mod 8).

. . . . . .

MA1100 Tutorial Tutorial 4: Proof

Schedule of Today

Review concepts Tutorial: 4.23(absolute value), 4.33(set relation), 4.42(set relation), 5.4(parity, disprove), 5.13(rational, irrational, proof by contradiction), 5.22(parity, existence, proof by contradiction), 5.32(existence, proof by contradiction), 5.33(existence, intermediate value theorem, uniqueness)

. . . . . .

MA1100 Tutorial Tutorial 4: Proof Review

Proof by Contradiction

Direct Proof, Proof by Contrapositive, Proof by contradiction; Prove R is true: Assume R is true, try to get a contradiction; Prove (@x)R(x) is true: Assume (Dx) R(x) is true, try to get a contradiction; Prove (@x)P(x) Q(x) is true: Assume (Dx)P(x) ^ ( Q(x)) is true, try to get a contradiction. When to use: When there is no direct proof: “there do not exist...”, “A is an emptyset”, “p is an irrational number”. When it is easy to work with the negation. Advantage: For implication P Q, we have more assumption4 to work with: For direct proof, we have one assumption P; For proof by contradiction, we have more assumption Q.

4Compare with disjunction in conclusion.

. . . . . .

MA1100 Tutorial Tutorial 4: Proof Review

Existence Proof

Three types: (Dx)P(x): (Dx)(@y)P(x, y): (@x)(Dy)P(x, y): Two approaches: Constructive proof: 1 Give a specific example of such objects; 2 Justify that the given examples satisfy the stated conditions. Non-constructive proof: 1 Use when specific examples are not easy or not possible to find; 2 Make arguments why such objects have to exist; 3 Use definitions, axioms or results that involves existence statements.

. . . . . .

MA1100 Tutorial Tutorial 4: Proof Review

Irrational Numbers

Rational A rational number is a real number that can be written as a quotient

m n where m and n are integers, with n 0.

Irrational An irrational number is a real number that is not a rational number. Properties Rational Rational = Rational, Rational Irrational = Irrational, Irrational Irrational = ?(it depends). Decimal Decimal      finite decimal, Rational non-terminating recurring decimal, Rational non-terminating non-recurring decimal, Irrational . Lowest Term A rational number m

n with n 0 is in lowest term if m and n have

no common factor which is greater than 1. This property is very useful for proof by contradiction.

. . . . . .

MA1100 Tutorial Tutorial 4: Proof Review

Sets Relations

Element-Chasing Method: 1 Choose an arbitrary element; 2 Show that the element satisfies the given property. 3 Using this method, we can prove: A B, A B, A = B, A B. Set Operations: P : x P A, Q : x P B,

Set Meaning Logic A X B x P A and x P B P ^ Q A Y B x P A or x P B P _ Q Ac x R A P A B a P A and x R B P ^ ( Q)

Algebra of Sets: Idempotent A X A = A, A Y A = A; Commutative A X B = B X A, A Y B = B Y A; Associative (A X B) X C = A X (B X C), (A Y B) Y C = A Y (B Y C); Distributive A X (B Y C) = (A X B) Y (A X C), A Y (B X C) = (A Y B) X (A Y C); Double Complement (Ac)c = A; De Morgan (A X B)c = Ac Y Bc, (A Y B)c = Ac X Bc;

. . . . . .

MA1100 Tutorial Tutorial 4: Proof Tutorial

Exercise (4.23) Let x, y P R. Prove that |xy| = |x| |y|. Method For these questions, based on definition of absolute value, we consider some cases. Recall Equivalent definitions: |x| = { x, if x 0; x, if x 0. = { x, if x 0; x, if x 0. =      x, if x 0; 0, if x = 0; x, if x 0. . Proof. Assume x 0 and y 0: then xy 0, and |xy| = xy, |x| = x, |y| = y. Therefore, |xy| = xy = |x| |y|. Assume x 0 and y 0: then xy 0, and |xy| = xy, |x| = x, |y| = y. Therefore, |xy| = xy = |x| |y|. Assume x 0 and y 0: then xy 0, and |xy| = xy, |x| = x, |y| = y. Therefore, |xy| = xy = |x| |y|. Assume x 0 and y 0: then xy 0, and |xy| = xy, |x| = x, |y| = y. Therefore, |xy| = xy = (x)(y) = |x| |y|.

. . . . . .

MA1100 Tutorial Tutorial 4: Proof Tutorial

Exercise (4.33) Let A and B be sets. Prove that A Y B = A X B if and only if A = B. Proof. There are two directions: “If” Assume A = B, then A Y B = A = A X B. “Only if” Assume A Y B = A X B, we want to show A = B. Direct Proof: Since A = B iff A B and B A, it suffices to show A B and B A. Using element-chasing method: for any x P A, then x P A Y B = A X B, therefore x P B, that is, A B; By symmetry, we also get B A. Proof by contrapositive: assume A B, that is A B or B A, we want to show that A Y B A X B: When A B: By definition, there exists a P A and a R B. Since a R B, we have a R A X B. On the other hand, a P A implies that a P A Y B. Therefore A Y B A X B. When B A: Similar.

. . . . . .

MA1100 Tutorial Tutorial 4: Proof Tutorial

Exercise (4.42) For sets A and B, fine a necessary and sufficient condition for (A B) X (B A) = H. Verify that this condition is necessary and sufficient. Recall Cartesian Product: A B = {(x, y) : x P A, y P B}. Method Start from assumption, we try to get some necessary(sufficient) condition, then try to prove the necessary(sufficient) condition is sufficient(necessary). Solution. .

.

. 1 Assume (A B) X (B A) H, then there exists (x, y) P (A B) X (B A), that is x P A, y P B and x P B, y P A, i.e. A X B H. (A X B = H is sufficient condition for original statement.) .

.

. 2 Guess: when A X B H, we have (A B) X (B A) H. (If not, we must try to add more conditions.) .

.

. 3 Assume A X B H, then there exists x P A X B, therefore (x, x) P (A B) X (B A), i.e. (A B) X (B A) H. .

.

. 4 Therefore, we find a necessary and sufficient condition: A X B = H.

. . . . . .

MA1100 Tutorial Tutorial 4: Proof Tutorial

Exercise (5.4) Disprove the statement: Let n P N. If n(n+1)

2

is odd, then (n+1)(n+2)

2

is odd. Method When disproving a statement, we only need a counterexample, and one counterexample is enough. For these questions, generally there are two methods: Enumerating: try 1, 2, 3, . . .; Start from assumption, to get some properties, then using the properties and enumerating method to find the number. Solution of Method 1. We want to find an integer n, such that n(n+1)

2

is odd and (n+1)(n+2)

2

is even.

n 1 2 3 4 5 6 7 8

n(n+1) 2

1 3 6 10 15 21 28 36

• dd
• dd

even even

• dd
• dd

even even

(n+1)(n+2) 2

3 6 10 15 21 28 36 45

• dd

even even

• dd
• dd

even even

• dd

Thus n = 2 is a counterexample.

. . . . . .

MA1100 Tutorial Tutorial 4: Proof Tutorial

Exercise (5.4) Disprove the statement: Let n P N. If n(n+1)

2

is odd, then (n+1)(n+2)

2

is odd. Solution of Method 2. Assume that n(n+1)

2

is odd and (n+1)(n+2)

2

is even, we want to find some properties

• f n.

.

.

. 1 Since (n+1)(n+2)

2

= n(n+1)

2

+ (n + 1), n(n+1)

2

is odd and (n+1)(n+2)

2

is even, we have n is even, say 2k, k P N; .

.

. 2 Substitute n = 2k to n(n+1)

2

and (n+1)(n+2)

2

: k(2k + 1) is odd and (k + 1)(2k + 1) is even, therefore k is odd, say 2p + 1, p 0; .

.

. 3 We get n = 2(2p + 1), p 0. .

.

. 4 We can use enumerating method for the restricted set {n P N : n = 2(2p + 1), p 0}. Letting p = 0, we get that n(n+1)

2

is odd, then

(n+1)(n+2) 2

is even when n = 2.

. . . . . .

MA1100 Tutorial Tutorial 4: Proof Tutorial

Exercise (5.13V) Prove that the product of an irrational number and a nonzero rational number is irrational. Recall R has two parts: rational numbers Q and irrational number R Q. Any rational number a can be expressed as p

q , where p, q P Z and q 0.

Method For these questions, in general, we use a proof by contradiction. Proof. .

.

. 1 Assume that there exist an irrational number a and a nonzero rational number b such that ab is rational. .

.

. 2 By definition, b can be expressed as r

s , where r, s P Z and r, s 0; ab can be

expressed as p

q , where p, q P Z and q 0.

.

.

. 3 Then a =

p bq = sp rq . Since sp, rq P Z and rq 0, it follows that a is a rational

number, which is a contradiction.

. . . . . .

MA1100 Tutorial Tutorial 4: Proof Tutorial

Exercise (5.22V) Let m be a positive integer of the form m = 2s, where s is an odd integer. Prove that there do not exist positive integers x and y such that x2 y2 = m. Method For these questions, it is difficult to give a direct proof, since the number of integers is infinite. Proof. Here we use a proof by contradiction: .

.

. 1 Assume that there exist positive integers x and y such that x2 y2 = m = 2s. Then (x + y)(x y) = 2s, where s is an odd integer. .

.

. 2 Generally we may consider four cases based on the parities of x and y. While, here we consider two cases, according to whether x and y are of the same parity

• r of opposite parity.

If x and y are of the same parity, then both x + y and x y are even, then 2s has factor 4, i.e. s has factor 2, it is a contradiction; If x and y are of opposite parity, then both x + y and x y are odd, therefore 2s is odd. Contradiction.

. . . . . .

MA1100 Tutorial Tutorial 4: Proof Tutorial

Exercise (5.32V) Show that there exist no nonzero real numbers a and b such that √ a2 + b2 =

3

√ a3 + b3. Method For these questions, it is difficult to give a direct proof, since the number of real numbers is infinite. Proof. Here we use a proof by contradiction: .

.

. 1 Assume that there exist nonzero real numbers a and b such that √ a2 + b2 =

3

√ a3 + b3. .

.

. 2 Raising both sides to the 6th power, we obtain a6 + 3a4b2 + 3a2b4 + b6 = a6 + 2a3b3 + b6. Thus 3a2 2ab + 3b2 = (a b)2 + 2a2 + 2b2 = 0. .

.

. 3 Since this can only occur when a = b = 05, we have a contradiction.

5Non-negative+Non-negative=0 iff each of them is zero.

. . . . . .

MA1100 Tutorial Tutorial 4: Proof Tutorial

Exercise (5.33) Prove that there exist a unique real number solution to the equation x3 + x2 1 = 0 between x = 2

3 and x = 1.

Method For existence of these questions, we will use Intermediate Value Theorem if we can not solve the equation directly. Recall Intermediate Value Theorem: If the function y = f(x) is continuous on the interval [a, b], and f(a)f(b) 0, then there is a c P [a, b] such that f(c) = 0. Proof of existence. Let f(x) = x3 + x2 1. .

.

. 1 Since f is a polynomial function, it is continuous on the set of all real numbers and so f is continuous on the interval [2/3, 1]. .

.

. 2 Because f(2/3) = 7/27 0 and f(1) = 1 0, it follows by the Intermediate Value Theorem that there is a number c between x = 2/3 and x = 1 such that f(c) = 0. Hence c is a solution.

. . . . . .

MA1100 Tutorial Tutorial 4: Proof Tutorial

Method For uniqueness, in general, let c1 and c2 be the numbers each of which satisfies the condition, and then prove c1 = c2. Proof of uniqueness. We now show that c is the unique solution of f(x) = 0 between 2/3 and 1. .

.

. 1 Let c1 and c2 be solutions of f(x) = 0 between 2/3 and 1. .

.

. 2 Then c3

1 + c2 1 1 = 0 and c3 2 + c2 2 1 = 0. Hence c3 1 + c2 1 1 = c3 2 + c2 2 1,

implying that c3

1 + c2 1 = c3 2 + c2 2 and so

c3

1 c3 2 + c2 1 c2 2 = (c1 c2)(c2 1 + c1c2 + c2 2) + (c1 c2)(c1 + c2)

= (c1 c2)(c2

1 + c1c2 + c2 2 + c1 + c2) = 0.

.

.

. 3 Since c2

1 + c1c2 + c2 2 + c1 + c2 0 (because 2/3 c1, c2 1), we obtain

c1 c2 = 0 and so c1 = c2.

. . . . . .

MA1100 Tutorial Mid-Term Exam

Time: Oct 2nd, 12-2pm; Venue: MPSH1; Close book with 1 helpsheet; Consultation:

Sept 28th, Monday, 13:00–17:00; Oct 1st, Thursday, 09:00–11:00, 13:00–17:00, 19:00–21:00; S9a-02-03.

Wiki for MA1100: http://wiki.nus.edu.sg/display/MA1100/MA1100+Home Results without proof: http://wiki.nus.edu.sg/display/MA1100/MA1100+Basic+Results

. . . . . .

MA1100 Tutorial Tutorial 5: Mathematical Induction

Schedule of Today

Review concepts Tutorial: 6.8, 6.9, 6.16, 6.20, 6.22, 6.35, 6.37, 6.51

. . . . . .

MA1100 Tutorial Tutorial 5: Mathematical Induction Review

Principle of Mathematical Induction

Axiom of Induction If T is a subset of N, such that: 1 P T; For every k P N, if k P T, then k + 1 P T. Then T = N. Well-Ordered Let H S R, S is well-ordered if every nonempty subset of S has smallest element. Well-Ordering Principle The set N is well-ordered. PMI Let P(n) be an open sentence, such that If P(1) is true; For all k P N, if P(k) is true, then P(k + 1) is true. Then P(n) is true for all n P N. SPMI Let P(n) be an open sentence, such that If P(1) is true; For all k P N, if P(1), P(2), . . . , P(k) are true, then P(k + 1) is true. Then P(n) is true for all n P N.

. . . . . .

MA1100 Tutorial Tutorial 5: Mathematical Induction Review

Applications and Generalizations

PMI(or SPMI) can be used to prove (@n P N)P(n) is true: Identify universal set and open sentence P(n); Base case: prove P(1) is true; Inductive step: for all k P N, if P(k)(or P(1) ^ P(2) ^ ^ P(k)) is true, prove P(k + 1) is true; Summary the conclusion you get. Generalization of universal set with some “good” order. {n : n P Z, n M} with the order: P(M) P(M+1) P(M+2) P(M+n) P(M+n+1) where M is an integer; Z with the order: P(n) P(2) P(1) P(0) P(1) P(2) P(n) ; Q+ with order in lecture notes; {n P N : n = 3p + 1, p P N} with the natural order: P(1) P(4) P(7) P(3p + 1) P(3p + 4) .

. . . . . .

MA1100 Tutorial Tutorial 5: Mathematical Induction Tutorial

Exercise (6.8) (a) We have seen that 12 + 22 + + n2 is the number of the squares in an n n square composed of n2 1 1 squares. What does 13 + 23 + 33 + + n3 represent geometrically? (b) Use mathematical induction to prove that 13 + 23 + 33 + + n3 = n2(n+1)2

4

for every positive integer n. Solution for (a). Let C be an n n n cube composed of n3 1 1 1 cubes. Then the number of different cubes that C contains is 13 + 23 + 33 + + n3.

. . . . . .

MA1100 Tutorial Tutorial 5: Mathematical Induction Tutorial

Proof for (b). Universal set {n : n P N}, and P(n) : 13 + 23 + + n3 = n2(n+1)2

4

. (1) Base case: when n = 1, LHS = 13 = 1 = 12(1+1)2

4

= RHS, i.e. P(1) is true; (2) Inductive step: for all k 1: Assume that P(k) is true, i.e. 13 + 23 + + k3 = k2(k+1)2

4

; Then we want to show that P(k + 1) is true, i.e. 13 + 23 + + (k + 1)3 = (k+1)2(k+1+1)2

4

: 13 + 23 + + (k + 1)3 = 13 + 23 + + k3

• k2(k+1)2/4

+(k + 1)3 = k2(k + 1)2 4 + (k + 1)3 = (k + 1)2(k2 + 4k + 4) 4 = (k + 1)2(k + 1 + 1)2 4 (3) By the Principle of Mathematical Induction, we have 13 + 23 + 33 + + n3 = n2(n+1)2

4

for every positive integer n.

. . . . . .

MA1100 Tutorial Tutorial 5: Mathematical Induction Tutorial

Exercise (6.9) Prove that 1 3 + 2 4 + 3 5 + + n(n + 2) = n(n+1)(2n+7)

6

for every positive integer n. Proof. Universal set is {n : n P N}, and P(n) : 1 3 + 2 4 + + n(n + 2) = n(n+1)(2n+7)

6

. (1) Base case: when n = 1, LHS = 1(1 + 2) = 3 = 129

6

= RHS, i.e. P(1) is true; (2) Inductive step: for all k 1: Assume that P(k) is true, i.e. 1 3 + 2 4 + + k(k + 2) = k(k+1)(2k+7)

6

; Then we want to show that P(k + 1) is true, i.e. 1 3 + 2 4 + + (k + 1)(k + 1 + 2) = (k+1)(k+1+1)(2(k+1)+7)

6

: 1 3 + 2 4 + + (k + 1)(k + 1 + 2) = 1 3 + 2 4 + + k(k + 2)

• k(k+1)(2k+7)/6

+(k + 1)(k + 1 + 2) = (k + 1)(k + 1 + 1)(2(k + 1) + 7) 6 (3) By the Principle of Mathematical Induction, we have 1 3 + 2 4 + 3 5 + + n(n + 2) = n(n+1)(2n+7)

6

for every positive integer n.

. . . . . .

MA1100 Tutorial Tutorial 5: Mathematical Induction Tutorial

Exercise (6.16,T) Prove that 1 + 1

4 + 1 9 + + 1 n2 2 1 n for every positive integer n.

Proof. Universal set is {n : n P N}, and P(n) : 1 + 1

4 + 1 9 + + 1 n2 2 1 n.

(1) Base case: when n = 1, LHS = 1 = 2 1 = RHS, i.e. P(1) is true; (2) Inductive step: for all k 1: Assume that P(k) is true, i.e. 1 + 1

4 + 1 9 + + 1 k2 2 1 k ;

Then we want to show 1 + 1

4 + 1 9 + + 1 (k+1)2 2 1 (k+1):

1 + 1 4 + + 1 (k + 1)2 = 1 + 1 4 + + 1 k2

• 2 1

k

+ 1 (k + 1)2 2 1 k + 1 (k + 1)2 = 2 k2 + k + 1 k(k + 1)2 2 k2 + k k(k + 1)2 = 2 1 k + 1 (3) By the Principle of Mathematical Induction, 1 + 1

4 + 1 9 + + 1 n2 2 1 n for

every positive integer n.

. . . . . .

MA1100 Tutorial Tutorial 5: Mathematical Induction Tutorial

Exercise (6.20,T) Prove that 7 (32n 2n) for every nonnegative integer n. Proof. Universal set is {n P Z : n 0}, and P(n) : 7 (32n 2n). (1) Base case: when n = 0, 30 20 = 0 and 7 0, i.e. P(0) is true; (2) Inductive step: for all k 0: Assume that P(k) is true, that is, 7 (32k 2k), i.e. 32k = 2k + 7a for some integer a; Then we want to show 7 (32(k+1) 2k+1), and it suffices to show that 32(k+1) 2k+1 has factor 7: 32(k+1) 2k+1 = 32 32k 2 2k = 9(2k + 7a) 2 2k = 7(2k + 9a). (3) By the Principle of Mathematical Induction, we have 7 (32n 2n) for every nonnegative integer n. Remark When n 1, by an bn = (a b)(an1 + an2b + + bn1), we have a direct proof: 32n 2n = 9n 2n = (9 2)c = 7c, where c = 9n12 + 9n222 + + 92n1 is an integer.

. . . . . .

MA1100 Tutorial Tutorial 5: Mathematical Induction Tutorial

Exercise (6.22,T) Prove that if A1, A2, . . . , An are any n 2 sets, then A1 X X An = A1 Y Y An. Proof. Universal set is {n P Z : n 2}, and P(n) : A1 X A2 X X An = A1 Y A2 Y Y An. (1) Base case: when n = 2, by De Morgan’s law, we have A1 X A2 = A1 Y A2 for any sets A1, A2, i.e. P(2) is true; (2) Inductive step: for all k 2: Assume that P(k) is true, i.e. A1 X X Ak = A1 Y Y Ak for any k sets A1, . . . , Ak; Then we want to show that P(k + 1) is true, i.e. A1 X X Ak+1 = A1 Y Y Ak+1 for any k + 1 sets A1, . . . , Ak+1: A1 X A2 X X Ak+1 = (A1 X X Ak) X Ak+1 Associated law = A1 X X Ak Y Ak+1 De Morgan’s law = ( A1 Y Y Ak ) Y Ak+1 Hypothesis = A1 Y Y Ak+1 Associated law (3) By the Principle of Mathematical Induction, we have A1 X X An = A1 Y Y An for any n sets.

. . . . . .

MA1100 Tutorial Tutorial 5: Mathematical Induction Tutorial

Exercise (6.35) Consider the sequence F1, F2, F3, . . ., where F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 = 5, and F6 = 8. The terms of this sequence are called Fibonacci numbers. (a) Define the sequence of Fibonacci numbers by means of a recurrence relation. (b) Prove that 2 Fn if and only if 3 n. Solution for (a). The sequence {Fn} is defined recursively by F1 = 1, F2 = 1, and Fn = Fn1 + Fn2 for n 3.

. . . . . .

MA1100 Tutorial Tutorial 5: Mathematical Induction Tutorial

Proof for (b) “if” case. If 3 n, we want to show 2 Fn. Universal set is {n : n = 3q, q P N} = {3, 6, . . . , 3k, 3(k + 1), . . .}, and P(n) : 2 Fn. (1) Base case: when n = 3 and F3 = 2, so 2 F3, i.e. P(3) is true. (2) Inductive step, for all n of form n = 3k with k P N: Assume that P(3k) is true, i.e. 2 F3k; Then we want to show that P(3(k + 1)) is true, i.e. 2 F3(k+1): By recurrent relation, we have F3(k+1) = F3k+2 + F3k+1 = F3k+1 + F3k + F3k+1 = 2F3k+1 + F3k which is even, i.e. 2 F3(k+1). (3) By the Principle of Mathematical Induction, we have that 2 Fn for all n which satisfies 3 n.

. . . . . .

MA1100 Tutorial Tutorial 5: Mathematical Induction Tutorial

Proof for (b) “only if” case. If 3 n, we want to show 2 Fn. Using prove by cases: n = 3q + 1 for q 0. Universal set is {n : n = 3q + 1, q 0, q P Z} = {1, 4, 7, . . . , 3k + 1, 3(k + 1) + 1, . . .}, Q(n) : 2 Fn. (1) Base case: when n = 1, F1 = 1, so 2 F1, i.e. Q(1) is true. (2) Inductive step, for all n of form n = 3k + 1 for k 0: Assume that Q(3k + 1) is true, i.e. 2 F3k+1; Then we want to show that Q(3(k + 1) + 1) is also true, i.e. 2 F3(k+1)+1: By recurrent relation, we have F3(k+1)+1 = F3k+3+F3k+2 = F3k+2+F3k+1+F3k+2 = 2F3k+2+F3k+1 which is odd, i.e. 2 F3(k+1)+1. (3) By the Principle of Mathematical Induction, we have that 2 Fn for all n which is of form n = 3q + 1. n = 3q + 2 for q 0. Universal set is {n : n = 3q + 2, q 0, q P Z} = {2, 5, 8, . . . , 3k + 2, 3(k + 1) + 2, . . .}, R(n) : 2 Fn. By similar method, we can prove 2 Fn for all n of form n = 3q + 2.

. . . . . .

MA1100 Tutorial Tutorial 5: Mathematical Induction Tutorial

Exercise (6.37) Use the Strong Principle of Mathematical Induction to prove that for each integer n 12, there are nonnegative integers a and b such that n = 3a + 7b. Proof. Universal set is {n P Z : n 12}, and P(n) : n = 3a + 7b for some integers a, b 0. (1) Base case: for n = 12, it is clear that 12 = 3 4, that is, we can choose a = 4 and b = 0, such that 12 = 3a + 7b, i.e. P(12) holds; (2) Inductive step: for all k 12: Assume that for every integer i with 12 i k, there exist integers a, b 0 such that i = 3a + 7b; We want to show that there exist integers x, y 0 such that k + 1 = 3x + 7y; Since 13 = 3 2 + 7 1 and 14 = 3 0 + 7 2, we may assume that k 14; Since k k 2 12, there exist integers c, d 0 such that k 2 = 3c + 7d; Hence k + 1 = 3(c + 1) + 7d. (3) By the Strong Principle of Mathematical Induction, for each integer n 12, there are integers a, b 0 such that n = 3a + 7b.

. . . . . .

MA1100 Tutorial Tutorial 5: Mathematical Induction Tutorial

Exercise (6.51) By an n-gon, we mean an n-sided polygon. So a 3-gon is a triangle and a 4-gon is a

• quadrilateral. It is well known that the sum of the interior angles of a triangle is 1800.

Use induction to prove that for every integer n 3, the sum of the interior angles of an n-gon is (n 2) 1800. Proof. For convenience, we use some notations: π = 1800, Sn denotes the sum of the interior angles of an n-gon. Universal set is {n P Z : n 3}, and P(n) : Sn = (n 2)π. (1) Base case: for n = 3, S3 = π = (3 2) 1800, i.e. P(3) holds; (2) Inductive step: for all k 3, .

.

. 1 Let Qk+1 be a (k + 1)-gon whose vertices are v1, v2, . . . , vk, vk+1 and whose edges are v1v2, v2v3, . . . , vkvk+1, vk+1v1; .

.

. 2 Then let Qk be the k-gon such that whose vertices are v1, v2, . . . , vk and whose edges are v1v2, v2v3, . . . , vkv1, Q3 be the 3-gon whose vertices are vk, vk+1, v1 and whose edges are vkvk+1, vk+1v1, v1vk; .

.

. 3 Observe that Sk+1 = Sk + S3; .

.

. 4 By the induction hypothesis, Sk is (k 2)π and S3 is π; .

.

. 5 Therefore, Sk+1 is (k 2)π + π = (k 1)π. (3) By Principle of Mathematical Induction, the sum of the interior angles of an n-gon is (n 2) 1800.

. . . . . .

MA1100 Tutorial Tutorial 6: Relations

Schedule of Today

Review concepts Tutorial: 8.6, 8.9, 8.10, 8.14, 8.17, 8.20, 8.23, 8.26

. . . . . .

MA1100 Tutorial Tutorial 6: Relations Review

Let A, B be sets. A relation R from A to B is a subset of A B, i.e. R = {(a, b) P A B : condition on a and b}. If (x, y) P R, then x is related to

• y. Notation: (x, y) P R, x y, x R y, xRy.

Let A be a set. A relation R on A is the subset of A A, i.e. R = {(a, b) P A A : condition on a and b}. Let R be a relation on A: R is reflexive on A if “for all x P A, xRx”; R is symmetric on A if “for all x, y P A, if xRy, then yRx”; R is transitive on A if “for all x, y, z P A, if xRy and yRz, then xRz”. Let R be a relation on A. R is an equivalence relation if it is a reflexive, symmetric, transitive relation on A. Let R be an equivalence relation on A. For each n P A, let [n]R = {x P A : (x, n) P R} = {x P A : (n, x) P R}. We call this an equivalence class of n determined by the relation R. Let R be an equivalence relation on A, then the collection C of all equivalence classes determined by R is a partition of the set A.

. . . . . .

MA1100 Tutorial Tutorial 6: Relations Review

Relation

Relation from A to B Relation on a Set A when B = A

❄

Reflexive Symmetric Transitive ✲ Equiv Relation R

✲ ✛

Equiv Classes {[n]R : n P A} Partition {[n]R : n P A} Thm 8.3

❄

Thm 8.4

✻ ✲ ✛

. . . . . .

MA1100 Tutorial Tutorial 6: Relations Tutorial

Exercise (8.6) Let S = {a, b, c}. Then R = {(a, a), (a, b), (a, c)} is a relation on S. Which of the properties reflexive, symmetric, and transitive does the relation R possess? Justify your answers. Solution. Not reflexive: (b, b) R R, (c, c) R R; Not symmetric: (a, b) P R but (b, a) R R, (a, c) P R but (c, a) R R; Transitive: The only ordered pairs (x, y) and (y, z) that belong to R are where (x, y) = (a, a), since y has only one choice a. The possible choices for (y, z) in R are (a, a), (a, b), and (a, c). In every case, (x, z) = (y, z) P R and so R is transitive.

. . . . . .

MA1100 Tutorial Tutorial 6: Relations Tutorial

Exercise (8.9T) A relation R is defined on Z by aRb if |a b| 2. Which of the properties reflexive, symmetric, and transitive does the relation R possess? Justify your answers. Solution. R = {(a, b) : a, b P Z, |a b| 2}. Reflexive: for all n P Z, since |n n| = 0 2, we have nRn; Symmetric: for all m, n P Z such that mRn, i.e |m n| 2, it can be rewritten as |n m| 2, then we have nRm; Not transitive: (0, 2) P R, and (2, 4) P R, but (0, 4) R R.

. . . . . .

MA1100 Tutorial Tutorial 6: Relations Tutorial

Exercise (8.10) Let A = {a, b, c, d}. How many relations defined on A are reflexive, symmetric, and transitive and contain the ordered pairs (a, b), (b, c), (c, d)? Solution. We focus on a: Since R is reflexive, we have (a, a) P R; By the assumption, we have (a, b) P R; By the assumption (a, b), (b, c) P R, since R is transitive, we have (a, c) P R; Similarly, we have (a, d) P R. Then [a]R = {a, b, c, d} = A. So the associated partition is {A}, and there is only one way for partition, that is, there is only one equivalence relation.

. . . . . .

MA1100 Tutorial Tutorial 6: Relations Tutorial

Exercise (8.14) Let R be an equivalence relation on A = {a, b, c, d, e, f, g} such that aRc, cRd, dRg, and bRf. If there are three distinct equivalence classes resulting from R, then determine these equivalence classes and determine all elements of R. Solution. R is an equivalence relation and aRc, cRd, dRg: a, c, d, g is in the same equivalence class, namely [a]R. Moreover, |[a]R| 4; Similarly, b, f is in the same equivalence class, namely [b]R. Moreover, |[b]R| 2; Similarly, e is in the equivalence class [e]R. Moreover, |[e]R| 1; Notice: Now we do not know whether [a]R, [b]R, [e]R are pairwise disjoint; Since |A| = 7, and there are three distinct equivalence classes, therefore [a]R, [b]R, [e]R are pairwise disjoint and the three distinct equivalence classes are just [a]R = {a, c, d, g}, [b]R = {b, f}, [e]R = {e}. Otherwise, you can find contradictions easily.

. . . . . .

MA1100 Tutorial Tutorial 6: Relations Tutorial

Exercise (8.17T) Let A = {1, 2, 3, 4, 5, 6}. The relation R = {(1, 1), (1, 5), (2, 2), (2, 3), (2, 6), (3, 2), (3, 3), (3, 6), (4, 4), (5, 1), (5, 5), (6, 2), (6, 3), (6, 6)} is an equivalence relation on A. Determine the distinct equivalence classes. Solution. The elements each of which has relation with 1 are 1 and 5, then the equivalence class of 1 is [1]R = {1, 5} = [5]R; The elements each of which has relation with 2 are 2, 3 and 6, then the equivalence class of 2 is [2]R = {2, 3, 6} = [3]R = [6]R; The element which has relation with 4 is just 4 itself, then the equivalence class

• f 4 is [4]R = {4}.

Therefore there are three distinct equivalence classes, namely {1, 5}, {2, 3, 6} and {4}.

. . . . . .

MA1100 Tutorial Tutorial 6: Relations Tutorial

Exercise (8.20) A relation R on a nonempty set A is defined to be circular if whenever xRy and yRz, then zRx for all x, y, z P A. Prove that a relation R on A is an equivalence relation if and only if R is circular and reflexive. Proof. There are two directions to prove: : Assume that R is an equivalence relation: it suffices to show that R is circular. For all x, y, z P A such that xRy and yRz, then xRz (by transitive), then zRx (by symmetric), therefore R is circular. : Assume that R is reflexive and circular: it suffices to show that R is symmetric and transitive. For all x, y P A such that xRy, also we have xRx, then we have yRx (by circular), therefore R is symmetric; For all x, y, z P A such that xRy and yRz, then zRx (by circular), then xRz (by symmetric), therefore R is transitive.

. . . . . .

MA1100 Tutorial Tutorial 6: Relations Tutorial

Exercise (8.23T) Let R be a relation defined on the set N by aRb if either a b or b a. Prove or disprove: R is an equivalence relation. Proof. We have 2R1 and 1R3, but 2

✚

• R3. So we have that R is not transitive, therefore R is

not an equivalence relation.

. . . . . .

MA1100 Tutorial Tutorial 6: Relations Tutorial

Exercise (8.26) (a) Prove that the intersection of two equivalence relations on a nonempty set is an equivalence relation. (b) Consider the equivalence relations R2 and R3 defined on Z by aR2b if a b (mod 2) and aR3b if a b (mod 3). By (a), R1 = R2 X R3 is an equivalence relation on Z. Determine the distinct equivalence classes in R1. Proof of (a). Suppose that R1 and R2 are two equivalence relations defined on a set S. Let R = R1 X R2. Let a P S. Since R1 and R2 are equivalence relations on S, it follows that (a, a) P R1 and (a, a) P R2. Thus (a, a) P R and so R is reflexive; For all a, b P S such that aRb, then (a, b) P R = R1 X R2. Thus (a, b) P R1 and (a, b) P R2. Since R1 and R2 are symmetric, (b, a) P R1 and (b, a) P R2. Thus (b, a) P R and so bRa. Hence R is symmetric; For all a, b, c P S such that aRb and bRc, then (a, b) P R1 and (a, b) P R2 and (b, c) P R1 and (b, c) P R2. Since R1 and R2 are transitive, (a, c) P R1 and (a, c) P R2. Thus (a, c) P R and so aRc. Therefore, R is transitive.

. . . . . .

MA1100 Tutorial Tutorial 6: Relations Tutorial

Solution of (b). Let a P Z. If x P [a]R1, then (x, a) P R1 and so (x, a) P R2 and (x, a) P R3. Therefore, x a (mod 2) and x a (mod 3). Hence x = a + 2m and x = a + 3n for some integers m and n. Hence 2m = 3n and so n is even. Thus n = 2k for some integer k, implying that x = a + 3n = a + 3(2k) = a + 6k and so x a = 6k. Hence x a (mod 6). Thus [a]R1 {x P Z : x a (mod 6)}. When x a (mod 6), obviously x a (mod 2) and x a (mod 3), i.e. (x, a) P R2 X R3 = R1, then x P [a]R1. Therefore [a]R1 = {x P Z : x a (mod 6)}. Therefore the distinct equivalence classes in R1 are: [0]R1 = {. . . , 12, 6, 0, 6, 12, . . .}, [1]R1 = {. . . , 11, 5, 1, 7, 13, . . .}, [2]R1 = {. . . , 10, 4, 2, 8, 14, . . .}, [3]R1 = {. . . , 9, 3, 3, 9, 15, . . .}, [4]R1 = {. . . , 8, 2, 4, 10, 16, . . .}, [5]R1 = {. . . , 7, 1, 5, 11, 17, . . .}.

. . . . . .

MA1100 Tutorial Tutorial 7: Relations and Functions

Schedule of Today

Review concepts Tutorial: 8.30, 8.31, 8.36, 8.33, 8.37, 8.38, 8.42, 9.4, 9.7

. . . . . .

MA1100 Tutorial Tutorial 7: Relations and Functions Review

Let A, B be sets. A relation R from A to B is a subset of A B, i.e. R = {(a, b) P A B : condition on a and b}. If (x, y) P R, then x is related to

• y. Notation: (x, y) P R, x y, x R y, xRy.

Let A be a set. A relation R on A is the subset of A A, i.e. R = {(a, b) P A A : condition on a and b}. Let R be a relation on A: R is reflexive on A if “for all x P A, xRx”; R is symmetric on A if “for all x, y P A, if xRy, then yRx”; R is transitive on A if “for all x, y, z P A, if xRy and yRz, then xRz”. Let R be a relation on A. R is an equivalence relation if it is a reflexive, symmetric, transitive relation on A. Let R be an equivalence relation on A. For each n P A, let [n]R = {x P A : (x, n) P R} = {x P A : (n, x) P R}. We call this an equivalence class of n determined by the relation R. Let R be an equivalence relation on A, then the collection C of all equivalence classes determined by R is a partition of the set A.

. . . . . .

MA1100 Tutorial Tutorial 7: Relations and Functions Review

Relation

Relation from A to B Relation on a Set A when B = A

❄

Reflexive Symmetric Transitive ✲ Equiv Relation R

✲ ✛

Equiv Classes {[n]R : n P A} Partition {[n]R : n P A} Thm 8.3

❄

Thm 8.4

✻ ✲ ✛

. . . . . .

MA1100 Tutorial Tutorial 7: Relations and Functions Review

a b (mod n) n (a b) n (b a) a b = nk for some integer k. Arithmetic on Zn: For [a]n, [c]n P Zn, define [a]n + [c]n = [a + c]n, [a]n [c]n = [ac]n. A function from a set A to a set B is a rule that associate with every element x

• f A exactly one element of the set B.

A function is also called a mapping. Notation: f : A B. The set A is called the domain of the function, and the set B is called the codomain of the function. If a P A, then the element of B that is associated with a is denoted f(a). f(a) is called the image of a under f, and a is called a preimage of f(a) under f.

. . . . . .

MA1100 Tutorial Tutorial 7: Relations and Functions Tutorial

Exercise (8.30) Let R be the relation defined on Z by aRb if a + b 0 (mod 3). Show that R is not an equivalence relation. Proof. R = {(a, b) : a, b P Z, a + b 0 (mod 3)}. It is obvious that R is a relation. Reflexibility: Since 1 1, the relation R is not reflexive. Symmetry: Since a + b 0 (mod 3) if and only if b + a 0 (mod 3), R is symmetric. Transitivity: Since 2 + 1 0 (mod 3), 1 + 5 0 (mod 3) and 2 + 5 1 (mod 3), let a = 2, b = 1, c = 5, then we have 2R1, 1R5 and 2R /5. Therefore R is not transitive. Therefore, R is not an equivalence relation.

. . . . . .

MA1100 Tutorial Tutorial 7: Relations and Functions Tutorial

Exercise (8.31) The relation R on Z defined by aRb if a2 b2 (mod 4) is known to be an equivalence relation. Determine the distinct equivalence classes. Solution. Since aRb if a2 b2 (mod 4), we consider the 4 cases: x = 4n, x = 4n + 1, x = 4n + 2 and x = 4n + 3: x 4n 4n+1 4n+2 4n+3 x2 16n2 16n2 + 8n + 1 16n2 + 16n + 4 16n2 + 24n + 9 x2 (mod 4) 1 1 x2 { 0 (mod 4), if x = 4n or x = 4n + 2; 1 (mod 4), if x = 4n + 1 or x = 4n + 3. Base on the above, there are 2 equivalence classes: {x = 4n or 4n + 2 : n P Z} = {0, 2, 4, . . .}, {x = 4n + 1 or 4n + 3 : n P Z} = {1, 3, 5, . . .}.

. . . . . .

MA1100 Tutorial Tutorial 7: Relations and Functions Tutorial

Exercise (8.36) Let R be the relation defined on Z by aRb if a2 b2 (mod 5). Prove that R is an equivalence relation, and determine the distinct equivalence classes. Recall a b (mod n) n (a b) n (b a) a b = nk for some integer k. Proof. Reflexibility: For all a P Z, since a2 a2 = 0 = 5 0, it follows that a2 a2, i.e.

• aRa. Hence R is reflexive.

Symmetry: For all a, b P Z such that aRb, i.e. a2 b2 (mod 5), then a2 b2 = 5k for some integer k. Then b2 a2 = 5 (k), i.e. b2 a2 (mod 5), so bRa. Hnece R is symmetric. Transitivity: For all a, b, c P Z such that aRb and bRc, then a2 b2 (mod 5) and b2 c2 (mod 5). Then a2 b2 = 5m and b2 c2 = 5n, so we have a2 c2 = 5(m + n) i.e. a2 c2 (mod 5). Therefore aRc and R is transitive.

. . . . . .

MA1100 Tutorial Tutorial 7: Relations and Functions Tutorial

Solution. Since aRb if a2 b2 (mod 5), we consider the 5 cases: x = 5n, x = 5n + 1, x = 5n + 2, x = 5n + 3 and x = 5n + 4:

x 5n 5n+1 5n+2 5n+3 5n+4 x2 25n2 25n2 + 10n + 1 25n2 + 20n + 4 25n2 + 30n + 9 25n2 + 40n + 16 x2 (mod 5) 1 4 4 1

x2      0 (mod 5), if x = 5n; 1 (mod 5), if x = 5n + 1 or x = 5n + 4, 4 (mod 5), if x = 5n + 3 or x = 5n + 3. Base on the above, there are 3 equivalence classes: {x = 5n : n P Z}, {x = 5n + 1, 5n + 4 : n P Z}, {x = 5n + 2, 5n + 3 : n P Z}.

. . . . . .

MA1100 Tutorial Tutorial 7: Relations and Functions Tutorial

Exercise (8.33) A relation R is defined on Z by aRb if 5a 2b (mod 3). Prove that R is an equivalence relation. Determine the distinct equivalence classes. Method 1. Since R is a relation, it suffices to show that R is reflexive, symmetric and transitive: Reflexibility: For all a P Z, since 5a 2a = 3a and 3|3a, so we have aRa and that R is reflexive. Symmetry: For all a, b P Z such that aRb, then 5a 2b (mod 3), i.e. 5a 2b = 3k for some integer k. Observe that 5b 2a = 3(a + b) (5a 2b) = 3(a + b k), i.e. 3|(5b 2a), so bRa and R is symmetric. Transitivity: For all a, b, c P Z such that aRb and bRc, then 5a 2b (mod 3) and 5b 2c (mod 3). So 5a 2b = 3x and 5b 2c = 3y for some integer x, y. Observe that 5a 2c = (5a 2b) + (5b 2c) 3b = 3(x + y b), i.e. 3|(5a 2c) and 5a 2c (mod 3). Therefore aRc and R is transitive.

. . . . . .

MA1100 Tutorial Tutorial 7: Relations and Functions Tutorial

Method 2. Claim: 5a 2b (mod 3) if and only if a b (mod 3). ‘’: Assume 5a 2b (mod 3), then 5a 2b = 3k for some integer k. Then 2(a b) = 3(k a). Since 2 does not have the factor 3, a b should have factor 3, i.e. 3|(a b). Hence a b (mod 3). ‘’: Assume a b (mod 3), then a b = 3n for some integer n. Observe that 5a 2b = 5(a b) + 3b = 3(5n + b), then 3|(5a 2b) and 5a 2b (mod 3). Base on the claim and that a b (mod 3) is an equivalence relation, we know that 5a 2b (mod 3) is also an equivalence relation. Solution. Since aRb if a b (mod 3), it is obvious that [0] = {. . . , 3, 0, 3, 6, . . .}, [1] = {. . . , 5, 2, 1, 4, . . .}, [2] = {. . . , 4, 1, 2, 5, . . .}.

. . . . . .

MA1100 Tutorial Tutorial 7: Relations and Functions Tutorial

Exercise (8.37) Construct the addition and multiplication tables in Z4 and Z5. Solution. For Z4: + [0] [1] [2] [3] [0] [0] [1] [2] [3] [1] [1] [2] [3] [0] [2] [2] [3] [0] [1] [3] [3] [4] [1] [2]

• [0]

[1] [2] [3] [0] [0] [0] [0] [0] [1] [0] [1] [2] [3] [2] [0] [2] [0] [2] [3] [0] [3] [2] [1] For Z5, you can find the tables in lecture notes.

. . . . . .

MA1100 Tutorial Tutorial 7: Relations and Functions Tutorial

Exercise (8.38) In Z8, express the following sums and products as [r], where 0 r 8. (a). [2] + [6]; (b). [2] [6]; (c). [13] + [138]; (d). [13] [138]. Solution. [2] + [6] = [8] = [0]; [2] [6] = [12] = [4]; [13] + [138] = [3] + [2] = [5]; [13] [138] = [3] [2] = [6];

. . . . . .

MA1100 Tutorial Tutorial 7: Relations and Functions Tutorial

Exercise (8.42) Prove or disprove: (a) There exists an integer a such that ab 0 (mod 3) for every integer b. (b) If a P Z, then ab 0 (mod 3) for every b P Z. (c) For every integer a, there exists an integer b such that ab 0 (mod 3). Proof. (a) Existential quantifier and Universal quantifier: True, consider a = 0 and b = 3 for example; (b) Universal quantifier and Universal quantifier: False, consider a = b = 1; (c) Universal and Existential quantifier: True, for a given a, let b = 0.

. . . . . .

MA1100 Tutorial Tutorial 7: Relations and Functions Tutorial

Exercise (9.4) For the given subset Ai of R and the relation Ri(1 i 3) form Ai to R, determine whether Ri is a function from Ai to R. (a) A1 = R, R1 = {(x, y) : x P A1, y = 4x 3} (b) A2 = [0, 8), R2 = {(x, y) : x P A2, (y + 2)2 = x} (c) A3 = R, R3 = {(x, y) : x P A3, (x + y)2 = 4} Recall In the definition of function, there is only one restriction: every element x of A exactly

• ne element of the set B.

Solution. (a) The relation R1 is a function from A1 to R. (b) The relation R2 is not a function from A2 to R. For example, both (9, 1) and (9, 5) belong to R2. (c) The relation R3 is not a function from A3 to R. For example, both (0, 2) and (0, 2) belong to R3.

. . . . . .

MA1100 Tutorial Tutorial 7: Relations and Functions Tutorial

Exercise (9.7) Let A = {1, 2, 3} and B = {x, y}. Determine BA. Solution. |BA| = |B||A| = 23 = 8. For 1, there are 2 choices; For 2, there are also 2 choices; Similarly for 3. Therefore BA = {f1, f2, f3, f4, f5, f6, f7, f8}, and f1 = {(1, x), (2, x), (3, x)}, f2 = {(1, x), (2, x), (3, y)}, f3 = {(1, x), (2, y), (3, x)}, f4 = {(1, x), (2, y), (3, y)}, f5 = {(1, y), (2, x), (3, x)}, f6 = {(1, y), (2, x), (3, y)}, f7 = {(1, y), (2, y), (3, x)}, f8 = {(1, y), (2, y), (3, y)}.

. . . . . .

MA1100 Tutorial Tutorial 8: Functions

Schedule of Today

Review concepts Tutorial: 9.15, 9.22, 9.24, 9.29, 9.35, 9.36, 9.40, 9.46

. . . . . .

MA1100 Tutorial Tutorial 8: Functions Review

Let f : A B and g : B C be functions. Injective or One-to-one: Original definition: @x, y P A, if x y, then f(x) f(y). Working definition: @x, y P A, if f(x) = f(y), then x = y. Negation: Dx, y P A such that x y and f(x) = f(y). Surjective or Onto: Original definition: @y P B, Dx P A such that y = f(x). Alternative definition: Range(f) =Codoman(f). Negation: Dy P B such that @x P A, y f(x). Or Range(f) Codomain(f). Bijective: f is both an injective and surjective function.

. . . . . .

MA1100 Tutorial Tutorial 8: Functions Review

Function: A relation f from a set A to a set B to be a function from A to B, if the following two conditions are satisfied: For each element a P A, there is an element b P B, such that (a, b) P f. If (a, b), (a, c) P f, then b = c. Well-defined: If a function f satisfies the following condition, it is called well-defined: If (a, b), (a, c) P f, then b = c. Maybe we are confused here since every function must be well-defined by the definition of function. However, there are situations though when the definition

• f a function f may make it unclear whether f is well-defined. This can often
• ccur when a function is defined on the set of equivalence classes of an

equivalence relation. In conclusion, when a question ask us to prove that a function f is well-defined, it ask us to prove that this rule or this relation is well-defined. That is to say, we need to show “if (a, b), (a, c) P f, then b = c”, while we can not use the fact that every function is always well-defined.

. . . . . .

MA1100 Tutorial Tutorial 8: Functions Review

Two functions f1 and f2 are equal if and only if Domain(f1) = Domain(f2) and f1(a) = f2(a) for every a P Domain(f). Notation: f1 = f2. Composition: Let f : A B, g : B C and h : C D be functions. The composition of f and g is the function g f : A

f

• B

g

• C defined by

(g f)(x) = g(f(x)). Associated law: h (g f) = (h g) f. f idA = f, idB f = f. If f and g are injective (or surjective), then g f is injective (or surjective). If g f is injective, then f is injective, and g may be not. If g f is surjective, then g is surjective, and f may be not. Inverse: Let f : A B be a bijection, for a P A and b P B, we define inverse function by f1(b) = a if f(a) = b. If the inverse function is defined, then f is bijection. If f and g are bijections, then (g f)1 = (f1) (g1).

. . . . . .

MA1100 Tutorial Tutorial 8: Functions Tutorial

Exercise (9.15) A function f : Z Z is defined by f(n) = 5n + 2. Determine whether f is (a) injective, (b) surjective. Proof. (a) Injective: Using working definition. Let m, n P Z. Suppose f(m) = f(n), i.e., 5m + 2 = 5n + 2, which gives 5m = 5n, and hence m = n. Therefore f is injective. (b) Not surjective: Suppose f(n) = 0. Then 5n + 2 = 0 which implies 5n = 2, which is impossible for any integer n. There is no n P Z such that f(n) = 0. Hence f is not surjective.

. . . . . .

MA1100 Tutorial Tutorial 8: Functions Tutorial

Exercise (9.22) Let f : Z5 Z5 be a function defined by f([a]) = [2a + 3]. (a) Show that f is well-defined. (b) Determine whether f is bijective. Proof. (a) Let [a], [b] P Z5 such that [a] = [b]. We want to show that f([a]) = f([b]), that is, [2a + 3] = [2b + 3]. Since [a] = [b], it follows that a b (mod 5). Then a b = 5k for some integer k and therefore (2a + 3) (2b + 3) = 2(a b) = 5 2k. So [2a + 3] = [2b + 3], i.e., f([a]) = f([b]). Therefore f is well-defined. (b) As we know, Z5 = {[0].[1], [2], [3], [4]}. Since f([0]) = [3], f([1]) = [0], f([2]) = [2], f([3]) = [4] and f([4]) = [1], it follows that f is one-to-one and onto and so f is bijective.

. . . . . .

MA1100 Tutorial Tutorial 8: Functions Tutorial

Exercise (9.24) Let A = [0, 1] denote the closed interval of real numbers between 0 and 1. Give an example of two different bijective functions f1 and f2 from A A, neither of which is the identity function. Solution. f1(x) = x2 on [0, 1]; f2(x) = √x on [0, 1]; f3(x) = 1 x on [0, 1].

. . . . . .

MA1100 Tutorial Tutorial 8: Functions Tutorial

Exercise (9.29) Prove or disprove the following: (a) If two functions f : A B and g : B C are both bijective, then g f : A C is bijective. (b) Let f : A B and g : B C be two functions. If g is onto, then g f : A C is

• nto.

(c) Let f : A B and g : B C be two functions. If g is one-to-one, then g f : A C is one-to-one. (d) There exist functions f : A B and g : B C such that f is not onto and g f : A C is onto. (e) There exist functions f : A B and g : B C such that f is not one-to-one and g f : A C is one-to-one.

. . . . . .

MA1100 Tutorial Tutorial 8: Functions Tutorial

Proof. (a) True. Corollary 9.8 in textbook; (b) False. Let A = {1, 2}, B = {a, b} and C = {x, y}; and let f : A B and g : B C be defined by f = {(1, a), (2, a)} and g = {(a, x), (b, y)}. Then g f = {(1, x), (2, x)}. Thus g is onto but g f is not. (c) False. Counterexample is same as (b). (d) True. Constructive proof: Let A = {1, 2}, B = {a, b, c} and C = {x, y}; and let f : A B and g : B C be defined by f = {(1, a), (2, b)} and g = {(a, x), (b, y), (c, y)}. Thus g f = {(1, x), (2, y)} is onto and f is not. (e) False. We show that for functions f : A B and g : B C, if f is not

• ne-to-one, then g f : A C is not one-to-one. Since f is not one-to-one, there

exist a, b P A such that a b and f(a) = f(b). Thus (g f)(a) = g(f(a)) = g(f(b)) = (g f)(b) and so g f is not one-to-one.

. . . . . .

MA1100 Tutorial Tutorial 8: Functions Tutorial

Exercise (9.35) Let the functions f : R R and g : R R be defined by f(x) = 2x + 3 and g(x) = 3x + 5. (a) Show that f is one-to-one and onto. (b) Show that g is one-to-one and onto. (c) Determine the composition function g f. (d) Determine the inverse functions f1 and g1. (e) Determine the inverse function (g f)1 of g f and the composition f1 g1. Proof. (a) Let f(a) = f(b), where a, b P R. Then 2a + 3 = 2b + 3. Adding -3 to both side and dividing by 2, we have a = b and so f is one-to-one. Let r P R, we want to find x P R, such that f(x) = r, i.e., 2x + 3 = r. Hence x = r3

2 . Therefore f is surjective.

(b) The proof is similar to that in (a).

. . . . . .

MA1100 Tutorial Tutorial 8: Functions Tutorial

Solution. (c) (g f)(x) = g(f(x)) = g(2x + 3) = 3(2x + 3) + 5 = 6x 4. (d) Let y = 2x + 3. Then x = y3

2 . So we have f1 : R R with f1(y) = y3 2 .

Let y = 5 3x. Then x = 5y

3 . So we have g1 : R R with g1(y) = 5y 3 .

(e) By part (d), we have f1 g1(y) = y+4

6 . Let y = 6x 4. Then x = y+4 6 .

So we have (g f)1 : R R with (g f)1(y) = y+4

6 . We have the result:

f1 g1 = (g f)1.

. . . . . .

MA1100 Tutorial Tutorial 8: Functions Tutorial

Exercise (9.36) Let A = R {1} and define f : A A by f(x) =

x x1 for all x P A.

(a) Prove that f is bijective. (b) Determine f1 (c) Determine f f f. Proof and Solution. .

.

. 1 Let a, b P A such that f(a) = f(b). So

a a1 = b b1 . Then a(b 1) = b(a 1).

This implies ab a = ba b, and hence a = b. Hence f is injective. Let c P A, we want to find a P A, such that f(a) = c, i.e., c =

a a1 . Then

a =

c c1 . Hence f is surjective.

.

.

. 2 Let y =

x x1 . So y(x 1) = x which implies yx y = x. Thus yx x = y and

so x(y 1) = y. This gives x =

y y1 . Hence f1 : A A with f1(y) = y y1 .

We observe that f = f1. .

.

. 3 f f f = f (f f1) = f idA = f where idA is the identity function on A.

. . . . . .

MA1100 Tutorial Tutorial 8: Functions Tutorial

Exercise (9.40) Let f : R R be the function defined by f(x) = x2 + 3x + 4. (a) Show that f is not injective. (b) Find all pairs r1, r2 of real numbers such that f(r1) = f(r2). (c) Show that f is not surjective. (d) Find the set S of all real numbers such that if s P S, then there us no real number x such that f(x) = s. (e) What well-known set is the set S in (d) related to?

. . . . . .

MA1100 Tutorial Tutorial 8: Functions Tutorial

Proof. (a-b) Let a, b P R such that f(a) = f(b). Thus a2 + 3a + 4 = b2 + 3b + 4. So a2 + 3a = b2 + 3b and a2 b2 + 3a 3b = (a + b)(a b) + 3(a b) = (a b)(a + b + 3) = 0. Therefore, a = b or a + b = 3. Moreover, f is not injective. (c) f(x) = x2 + 3x + 4 = (x + 3

2 )2 + 7 4 7 4 . Thus there is no x P R such that

f(x) = 0 and so f is not surjective. (d) S = {s P R : s 7

4 }.

(e) This is the complement of the range of f.

. . . . . .

MA1100 Tutorial Tutorial 8: Functions Tutorial

Exercise (9.46) For each of the following functions, determine, with explanation, whether the function is one-to-one and whether it is onto. (a) f : R R R R, where f(x, y) = (3x 2, 5y + 7) (b) g : Z Z Z Z, where g(m, n) = (n + 6, 2 m) (c) h : Z Z Z Z, where h(r, s) = (2r + 1, 4s + 3) (d) φ : Z Z S = {a + b √ 2 : a, b P Z}, where φ(a, b) = a + b √ 2 (e) α : R R R, where α(x) = (x2, 2x + 1)

. . . . . .

MA1100 Tutorial Tutorial 8: Functions Tutorial

Solution. (a) Just checking definition, we get that f is one-to-one and onto. (b) Let (a, b), (c, d) P Z Z such that g(a, b) = g(c, d). Thus (b + 6, 2 a) = (d + 6, 2 c). So a = b and c = d. Therefore g is one-to-one. Let (c, d) P Z Z, a = 2 d and b = c 6. Then g(a, b) = (c, d). Hence g is surjective. (c) h is one-to-one by similar method of (b). While h is not onto, here is a counterexample: for (0, 0) P Z Z, if (2r + 1, 4s + 3) = (0, 0), then r = 1/2, s = 3/4 R Z. (d) It is obvious that φ is onto. Let (a, b), (c, d) P Z Z such that φ(a, b) = φ(c, d). Thus a + b √ 2 = c + d √

• 2. So a c = (d b)

√

• 2. Therefore a = c and b = d,

that is, φ one-to-one. (e) Let x, y P R such that α(x) = α(y). Thus (x2, 2x + 1) = (y2, 2y + 1). So x = y. Therefore α one-to-one. Since x2 0, α is not onto.

. . . . . .

MA1100 Tutorial Tutorial 9: Number Theory

Schedule of Today

Review concepts Tutorial: 11.8, 11.9, 11.18, 11.21, 11.30, 11.32, 11.42

. . . . . .

MA1100 Tutorial Tutorial 9: Number Theory Review

Prime: A prime is an integer p 2 whose only positive integer divisors are 1 and p. An integer n 2 that is not prime is called a composite number. Common divisor: Let a, b be integers and d a nonzero integer. d is a common divisor of a and b if d a and d b. Original definition: Let a, b be integers, not both 0. The largest integer that divides both a and b is called the greatest common divisor of a and

• b. gcd(0, 0) is not defined. Notation: gcd(a, b).

Working definition: Let a, b be integers, not both 0, and d P N. d = gcd(a, b) if and only if { d a and d b; For all k P N, if k a, k b, then k d. Theorem: Let a, b be integers, not both 0, and d P N. d = gcd(a, b) if and only if { d a and d b; For all k P N, if k a, k b, then k d. Division Algorithm: Original: For all positive integers a and b, there exist unique integers q and r, such that b = aq + r, where 0 r a. Generalization: For all integers a and b, there exist unique integers q and r, such that b = aq + r, where 0 r |a|. Here allow a and b to be negative.

. . . . . .

MA1100 Tutorial Tutorial 9: Number Theory Review

Euclidean Algorithm

Let a and b be positive integers. If b = aq + r for some integers q and r, then gcd(b, a) = gcd(a, r). Let a, b be integers, not both 0. b = a q1 + r1 gcd(b, a) = gcd(a, r1) 0 r1 |a| a = r1 q2 + r2 gcd(a, r1) = gcd(r1, r2) 0 r2 r1 r1 = r2 q3 + r3 gcd(r1, r2) = gcd(r2, r3) 0 r3 r2 rn3 = rn2 qn1 + rn1 gcd(rn3, rn2) = gcd(rn2, rn1) 0 rn1 rn2 rn2 = rn1 qn + rn gcd(rn2, rn1) = gcd(rn1, rn) 0 rn rn1 rn1 = rn qn+1 + 0 gcd(rn1, rn) = gcd(rn, 0) = rn ri will become 0 eventually since |a| r1 r2 rn1 rn 0. Then gcd(a, b) = rn.

. . . . . .

MA1100 Tutorial Tutorial 9: Number Theory Review

Linear combination: Given integers a and b. An integer n is called a linear combination of a and b if n can be written in the form ax + by by some integers x and y. a, b not both 0, then gcd(a, b) is the smallest positive linear combination

• f a and b.

Relatively Prime or Co-Prime: Let a and b be integers, not both 0. If gcd(a, b) = 1, then a and b are relatively prime. a and b are relatively prime if and only if 1 is a linear combination of a and b. Let a, b, c P Z, where a and b are relatively prime nonzero integers. If a c and b c, then ab c. Euclid’s Lemma Let a, b be integers, and p be a prime number. If p ab, then p a or p b. Let a1, a2, . . . , an be integers, and p be a prime number. If p a1 an, then p ak for some 1 k n.

. . . . . .

MA1100 Tutorial Tutorial 9: Number Theory Review

Prime factorization: n = p1p2 pr with primes p1 p2 . . . pr. We call this a prime factorization of n. Fundamental Theorem of Arithmetic: Existence of prime factorization: Every integer greater than 1 is either a prime number or a product of prime numbers. Uniqueness of prime factorization: For any integer greater than 1, the prime factorization is unique. Canonical factorization: Given any integer n 1. Suppose p1 p2 pr are the distinct prime divisors of n. Then we can write n = pk1

1 pk2 2 pkr r

for ki 1.

. . . . . .

MA1100 Tutorial Tutorial 9: Number Theory Tutorial

Exercise (11.8) Prove that 5 (33n+1 + 2n+1) for every positive integer n. Proof. We will use mathematical induction. The universal set is N. Base case: For n = 1, we have 33n+1 + 2n+1 = 34 + 22 = 85 and 5 85. Thus the result is true for n = 1. Inductive step: For all positive integer k, assume that 5 (33k+1 + 2k+1) for some positive integer k. We show that 5 (33(k+1)+1 + 2k+2). Since 5 (33k+1 + 2k+1), it follows that 33k+1 + 2k+1 = 5a for some integer a. Thus 33k+1 = 5a 2k+1. Now observe that 33(k+1)+1 + 2k+2 = 33 33k+1 + 2k+2 = 27(5a 2k+1) + 2k+2 = 5(27a) 25 2k+1 = 5(27a 5 2k+1). Since 27a 5 2k+1 is an integer, 5 (33(k+1)+1 + 2k+2). The result follows by the Principle of Mathematical Induction.

. . . . . .

MA1100 Tutorial Tutorial 9: Number Theory Tutorial

Exercise (11.9) Prove that for every positive integer n, there exist n consecutive positive integers, each of which is composite. Solution. Consider the n numbers: (n + 1)! + 2, (n + 1)! + 3, . . . , (n + 1)! + (n + 1). Observe for 2 k n + 1 that k divides k + (n + 1)!. Thus these n numbers are composite. Remark This result implies that we can find two consecutive primes which are as far apart as we want.

. . . . . .

MA1100 Tutorial Tutorial 9: Number Theory Tutorial

Exercise (11.18) (a) Prove that for every integer m, one of the integers m, m + 4, m + 8, m + 12, m + 16 is a multiple of 5. (b) State and prove a generalization of the result in (a). Proof. (a) Observe that m = 5q + r, where q, r P Z and 0 r 4. If m = 5q, then m is a multiple of 5. If m = 5q + 1, then m + 4 is a multiple of 5. If m = 5q + 2, then m + 8 is a multiple of 5. If m = 5q + 3, then m + 12 is a multiple of 5. If m = 5q + 4, then m + 16 is a multiple of 5.

. . . . . .

MA1100 Tutorial Tutorial 9: Number Theory Tutorial

Solution. Result: Let n P Z. For every integer m, one of the integers m, m + (n 1), m + 2(n 1), . . . , m + (n 1)2 is a multiple of n. Proof: By the Division Algorithm, there exist integers q and r such that m = nq + r, where 0 r n 1. For the number m + r(n 1), we have m + r(n 1) = (nq + r) + r(n 1) = nq + rn = n(q + r). Since q + r P Z, it follows that n [m + r(n 1)].

. . . . . .

MA1100 Tutorial Tutorial 9: Number Theory Tutorial

Exercise (11.21) (a) Prove that if an integer n has the form 6q + 5 for some q P Z, then n has the form 3k + 2 for some k P Z. (b) Is the converse of (a) true? Proof. (a) Observe that n = 6q + 5 = 3(2q) + 3 + 2 = 3(2q) + 1 + 2. Letting k = 2q + 1, we see that n = 3k + 2. (b) The converse is false. The integer 2 = 3 0 + 2 is of the form 3k + 2, but 2 is not

• f the form 6q + 5 since 6q + 5 = 2(3q + 2) + 1 is always odd.

. . . . . .

MA1100 Tutorial Tutorial 9: Number Theory Tutorial

Exercise (11.30) An integer n 1 has the properties that n (35m + 26) and n (7m + 3) for some integer m. What is n? Solution. Since n (7m + 3), it follows that n 5(7m + 3). Hence n [(35m + 26) 5(7m + 3)] = 11. Thus n = 11.

. . . . . .

MA1100 Tutorial Tutorial 9: Number Theory Tutorial

Exercise (11.32) Prove that following: Let a, b, c, m, n P Z, where m, n 2. If a b (mod m), a c (mod n), and d = gcd(m, n), then b c (mod d). Proof. Since a b (mod m), a b = mx for some integer x. Since a c (mod n), a c = ny for some integer y. Since d = gcd(m, n), d m and d n. Thus m = dr and n = ds for some integers r, s. b c = (a mx) (a ny) = (a drx) (a dsy) = d(sy rx) Therefore d (b c), i.e., b c (mod d).

. . . . . .

MA1100 Tutorial Tutorial 9: Number Theory Tutorial

Exercise (11.42) Prove the following: Let a, b, m, n P Z, where m, n 2. If a b (mod m), a b (mod n), and gcd(m, n) = 1, the a b (mod mn). Proof. Since a b (mod m) and a b (mod n), we have m (a b) and n (a b). Since gcd(m, n) = 1, by Theorem 11.16, mn (a b). Hence a b (mod mn).

. . . . . .

MA1100 Tutorial Tutorial 10: Number Theory and Cardinalities

Schedule of Today

Review concepts Tutorial: 11.46, 11.47, 11.48, 11.53, 10.12, 10.15, 10.18, 10.19

. . . . . .

MA1100 Tutorial Tutorial 10: Number Theory and Cardinalities Review

If there exists a bijective f : A B, we say that A is numerically equivalent to B. If A and B are two finite sets, then A is numerically equivalent to B if and only if |A| = |B|. If A and B are two infinite sets, we define |A| = |B| if A is numerically equivalent to B. If |A| = |N|, then A is called a denumerable set, and write |A| = |N| = ℵ0. A set is called a countable set if it is either finite or a denumerable set. A set that is not countable is called an uncountable set. If there exists a injective function f : A B but no bijective function, we say that A has smaller cardinality than B, write |A| |B|. From this, we can get: If there exists a injective function f : A B, then |A| |B|. Sets          finite infinite      denumerable: ℵ0 (N, Z, Q) uncountable { equivalent to (0, 1) : ℵ1, c ((a, b), [0, 1], [a, b], R, I) not equivalent to (0, 1) : ℵ2, ℵ3, . . . (P(R))

. . . . . .

MA1100 Tutorial Tutorial 10: Number Theory and Cardinalities Tutorial

Exercise (11.46) Prove each of the following: (a) Every prime of the form 3n + 1 is also of the form 6k + 1. (b) If n is positive integer of the form 3k + 2, then n has a prime factor of this form as well. Proof. (a) Let 3n + 1 be a prime. We claim that n must be even. If n is odd, then n = 2k + 1 for some integer k. So 3n + 1 = 3(2k + 1) + 1 = 6k + 4 = 2(3k + 2) where 3k + 2 2. Thus 3n + 1 is composite, which is impossible. Thus, as claimed, n is even and so n = 2k for some integer k. Therefore, 3n + 1 = 3(2k) + 1 = 6k + 1. (b) Let n be a positive integer such that n = 3l + 2, where l P Z. If n is a prime, then the proof is complete. Assume, to the contrary, that no prime factor of n is of the form 3k + 2 for some k P Z. We consider two cases. Some prime factor p of n is of the form 3k, where k P Z. Necessarily then, 3|p and so p = 3, contradicting our assumption that n = 3l + 2, where l P Z. Every prime factor of n is of the form 3k + 1, where k P Z. By Exercise 11.22, n is of the form 3k + 1, which is a contradiction.

. . . . . .

MA1100 Tutorial Tutorial 10: Number Theory and Cardinalities Tutorial

Exercise (11.47) (a) Express each of the integers 4278 and 71929 as a product of primes. (b) What is gcd(4278, 71929)? Recall

• pp. 258 in the textbook.

2|n iff the last digit of n is even; 3|n iff the sum of its digits is divisible by 3; 5|n iff the last digit of n is 0 or 5; Start with the first digit of n and sum alternate digits. Let the sum be a, b be the summation of the remaining digits. Then 11|n iff 11|(a b). Proof. (a) 4278 = 2 3 23 31 and 71929 = 11 13 503. (b) gcd(4278, 71929) = 1.

. . . . . .

MA1100 Tutorial Tutorial 10: Number Theory and Cardinalities Tutorial

Exercise (11.48) Let k be a positive integer. (a) Prove that if 2k 1 is prime, then k is prime. (b) Prove that if 2k 1 is prime, then n = 2k1(2k 1) is perfect. Proof. (a) Assume that k is not prime. If k = 1, then 2k 1 = 1 is not prime (contradiction). If k is a composite number, then k = ab, where a, b P Z and 1 a, b k. Therefore, 2k 1 = 2ab 1 = (2a)b 1. Letting x = 2a, we have 2k 1 = xb 1, where x 4. Since b 2, we have xb 1 = (x 1)(xb1 + xb2 + + 1), and x 1 3, xb1 + + 1 5. Thus (x 1)|(xb 1) and so 2k 1 is not prime. (b) Assume that 2k 1 is prime. Let p = 2k 1. Then k 2. The proper divisors of n = 2k1(2k 1) = 2k1p are then p, 2p, 22p, . . . , 2k2p and 1, 2, 22, . . . , 2k1. The sum of these integers is p(1 + 2 + 22 + + 2k2) + (1 + 2 + 22 + + 2k1) = p(2k1 1) + (2k 1) = (2k 1)[(2k1 1) + 1] = 2k1(2k 1) = n, as desired.

. . . . . .

MA1100 Tutorial Tutorial 10: Number Theory and Cardinalities Tutorial

Exercise (11.53) Evaluate the proposed solution of the following problem. Prove or disprove the following statement: There do not exist three integers n, n + 2, and n + 4, all of which are primes. Solution: This statement is true. Proof: Assume, to the contrary, that there exist three integers n, n + 2, and n + 4, all

• f which are primes. We can write n as 3q, 3q + 1, or 3q + 2, where q P Z. We

consider these three cases. Case 1 n = 3q. Then 3|n and so n is not prime. This is a contradiction. Case 2 n = 3q + 1. Then n + 2 = 3q + 3 = 3(q + 1). Since q + 1 is an integer, 3|(n + 2) and so n + 2 is not prime. Again, we have a contradiction. Case 3 n = 3q + 2. Hence we have n + 4 = 3q + 6 = 3(q + 2). Since q + 2 is an integer, 3|(n + 4). This produces a contradiction. Solution. It is wrong to say: “Then 3|n and so n is not prime.” Note that 3|3 and 3 is prime.

. . . . . .

MA1100 Tutorial Tutorial 10: Number Theory and Cardinalities Tutorial

Exercise (10.12) Let A = {a1, a2, a3, . . .}. Define B = A {an2 : n P N}. Prove that |A| = |B|. Proof. We need to check the condition of Theorem 10.3 .

.

. 1 Since we can find a bijective function f : A N, an n, A is denumerable. .

.

. 2 It is obvious that B is a subset of A. .

.

. 3 There are 2n numbers between an2 and a(n+1)2 in B, therefore the number of B is #B = 2 + 4 + 6 + =

8

∑

n=1

2n = 8. Hence, B is infinite. Therefore it follows that B is denumerable by Theorem 10.3.

. . . . . .

MA1100 Tutorial Tutorial 10: Number Theory and Cardinalities Tutorial

Exercise (10.15) Prove that the set of irrational numbers is uncountable. Proof. Denote the set of irrational numbers by I. Assume, to the contrary, that I is

• denumerable. Since Q and I are disjoint denumerable sets, Q Y I is denumerable by

Exercise 10.1. Since Q Y I = R, it follows that R is denumerable, which is a contradiction.

. . . . . .

MA1100 Tutorial Tutorial 10: Number Theory and Cardinalities Tutorial

Exercise (10.18) (a) Prove that the function f : (0, 1) (0, 2), mapping the open interval (0, 1) into the open interval (0, 2) and defined by f(x) = 2x, is bijective. (b) Explain why (0, 1) and (0, 2) have the same cardinality. (c) Let a, b P R, where a b. Prove that (0, 1) and (a, b) have the same cardinality. Proof. (a) Assume that f(a) = f(b), where a, b P (0, 1). Then 2a = 2b and so a = b. Hence f is one-to-one. For each r P (0, 2), x = r

2 P (0, 1) and f(x) = r. Therefore, f is

• nto. Thus f is a bijective function from (0, 1) to (0, 2).

(b) It follows by (a). (c) Define the function g : (0, 1) (a, b) by g(x) = (b a)x + a. Then g is bijective and so (0, 1) and (a, b) have the same cardinality.

. . . . . .

MA1100 Tutorial Tutorial 10: Number Theory and Cardinalities Tutorial

Exercise (Extra) Construct a bijective function f : [0, 1] (0, 1). Solution. f =     

1 2 ,

when x = 0;

1 n+2 ,

when x = 1

n, n P N;

x, when x 0 and is not reciprocal of some positive integer. Remark From this, we have: |[0, 1]| = |(0, 1)| = |[0, 1)| = |(0, 1]| = |[a, b]| = |(a, b)| = |[a, b)| = |(a, b]| = R.

. . . . . .

MA1100 Tutorial Tutorial 10: Number Theory and Cardinalities Tutorial

Exercise (10.19) Prove or disprove the following: (a) If A is an uncountable set, then |A| = |R|. (b) There exists a bijective function f : Q R. (c) If A, B and C are sets such that A B C, and A and C are denumerable, then B is denumerable. (d) The set S = {

√ 2 n

: n P N} is denumerable. (e) There exists a denumerable subset of the set irrational numbers. (f) Every infinite set is a subset of some denumerable set. (g) If A and B are sets with the property that there exists an injective function f : A B, then |A| = |B|.

. . . . . .

MA1100 Tutorial Tutorial 10: Number Theory and Cardinalities Tutorial

Proof. (a) False. For example, |P(R)| |R|. (b) False. |Q| |R|. (c) True. Since A is denumerable and A B, the set B is infinite. Since B is an infinite subset of the denumerable set C, it follows that B is denumerable. (d) True. Consider the function f : N S defined by f(n) =

√ 2 n . The function f is

bijective. (e) True. (See (d).) (f) False. Consider R. (g) False. The function f : N R defined by f(n) = n is injective but |N| |R|.

. . . . . .

MA1100 Tutorial Final Exam

Final Exam Information

Read Exam Info.pdf and Exam Seat Plan.pdf in Workbin on IVLE. Consultation: Any time from 14 Nov to 24 Nov, at my office S9a-02-03. Email: xiangsun@nus.edu.sg. MSN: xiangsun.sunny@hotmail.com. Read Lecture 24.pdf. Results available in final exam (from VT): Short answer: you can use all of them (that we have discussed and prove) Longer answer: you can use the results relative to the question asked. If an exam question can be answered in one or two lines by quoting a result, then you should know that you need to elaborate more. Wiki for MA1100: http://wiki.nus.edu.sg/display/MA1100/MA1100+Home. Be careful, and do not make any foolish mistakes. Good Luck.

. . . . . .

MA1100 Tutorial Thank you

Thank you